Equilibrium of a body: forces and moments
Why can a paint can be placed on the overhang of a plank lying across trestles, but only up to a certain point before the plank topples?
Prior knowledge
- The moment of a force is $M = Fr$, where $r$ is the lever arm of the force, and the sign of the moment tells the direction of rotationchapter Moment: the turning effect of a force
- Only the forces acting on the body are drawn in a free-body diagram, and for a body at rest the sum of the forces acting on it is zerochapter Free-body diagrams and the combined effect of forces
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
A painter has laid a long plank across two trestles so that one end of the plank reaches well beyond a trestle. He lifts a heavy paint can onto the plank and pushes it towards the end of the overhang. At last the other end of the plank lifts off its trestle and the plank topples.
What decides how far along the overhang the can can be moved, and how large a share of the load does each trestle carry?
Optional Recall: the moment of a force
Predict
Make your predictions before investigating 2 questions
Prediction 1: which support carries more
Prediction 2: when the plank topples
Investigate
Move the supports and the box. Follow the free-body diagram to see how the support forces change, and the situation picture to see where the common centre of mass of the plank and the box is.
Simulation
How to use
The sliders move supports 1 and 2 and the box; the positions are measured from the left end of the plank. Choose the mass of the box or no box. The dashed line of the situation picture passes through the common centre of mass of the plank and the box down to the metre scale on the floor; the ball at its lower end is green when the centre of mass is between the supports and red when it is not. Below is the free-body diagram of the plank on the same horizontal scale: $\overline{G}$ is the weight of the plank, $\overline{F}_\text{b}$ the force of the box on the plank, and $\overline{N}_1$, $\overline{N}_2$ the support forces. For a given mass of the box the arrows have the same scale at all settings; a heavier box makes the scale smaller so that even the largest support force fits in the picture. Reset restores the initial settings. The simulation has no animation, so no speed control is needed.
Assumptions of the model
The plank is uniform and rigid: length $4.0\,\text{m}$, mass $20\,\text{kg}$, centre of mass in the middle at $2.0\,\text{m}$. The supports are points that can only push the plank upwards. The box is at rest on the plank, so the force it exerts on the plank is equal to the weight of the box. When equilibrium is not possible, the plank is drawn toppled, and a negative support force in the table means that the support would have to pull the plank down, which a support cannot do. The values are calculated by the model, not measured.
How the calculation works
The weight is $G = m_\text{p} g$ and the force of the box $F_\text{b} = m_\text{b} g$, $g = 9.81\,\text{m/s}^2$. The moment condition about support 1, with anticlockwise positive:
$$\begin{aligned} &N_2 (x_2 - x_1) \\ &\quad - G (2.0\,\text{m} - x_1) \\ &\quad - F_\text{b} (x_\text{b} - x_1) = 0 \end{aligned}$$
from which $N_2$ is solved. Then from the sum of the forces $N_1 = G + F_\text{b} - N_2$. The readout shows the moments about support 1, and their sum is zero. The common centre of mass is $x_\text{cm} = \dfrac{m_\text{p} \cdot 2.0\,\text{m} + m_\text{b} x_\text{b}}{m_\text{p} + m_\text{b}}$.
Experiment 1: a support closer to the centre of mass
Test prediction 1. There is no box, and support 1 is at $0.5\,\text{m}$. Start with support 2 at $3.5\,\text{m}$ and move it towards the middle of the plank. Record the support forces at no fewer than three positions of support 2, for example $3.5\,\text{m}$, $3.0\,\text{m}$ and $2.5\,\text{m}$. Also calculate $N_1 + N_2$ on each row.
Initial state: support 1 = $0{.}5$, support 2 = $3{.}5$, box = $1$, mass of the box: no box. Kept constant: support 1, box, mass of the box. Varied: support 2.
| support 2 | $N_1$ $(\text{N})$ | $N_2$ $(\text{N})$ |
|---|
Experiment 2: the box onto the overhang
Test prediction 2. The supports are at $0.5\,\text{m}$ and $3.0\,\text{m}$, and the plank carries a $40\,\text{kg}$ box. Move the box to the right and record the support forces and the common centre of mass $x_\text{cm}$ at, for example, $3.0\,\text{m}$, $3.2\,\text{m}$, $3.4\,\text{m}$, $3.5\,\text{m}$ and $3.6\,\text{m}$. At which position is $N_1$ zero, and where is $x_\text{cm}$ then?
Initial state: support 1 = $0{.}5$, support 2 = $3$, box = $2$, mass of the box = $40$. Kept constant: support 1, support 2, mass of the box. Varied: box.
| box | $N_1$ $(\text{N})$ | $N_2$ $(\text{N})$ | $x_\text{cm}$ $(\text{m})$ | topples? |
|---|
Conclusion: support forces and toppling
On the basis of your tables, write: which support carried more, and how did the ratio of the support forces depend on the distances of the supports from the centre of mass of the plank? What stayed the same when you moved a support? Where was the box when the plank started to topple, and where was the common centre of mass then?
Model answer
In experiment 1 the support nearer the centre of mass carried more. When support 2 was at $2.5\,\text{m}$, that is $0.5\,\text{m}$ from the centre of mass, and support 1 $1.5\,\text{m}$ from it, the support forces were $N_2 = 147\,\text{N}$ and $N_1 = 49\,\text{N}$. The ratio $3 : 1$ is the inverse ratio of the distances, as the moment condition about the centre of mass, $N_1 r_1 = N_2 r_2$, requires. The sum of the support forces stayed equal to the weight of the plank, $196\,\text{N}$, because the sum of the forces is zero. In experiment 2 the plank still rested on the supports when the box was past support 2 at $3.2\,\text{m}$. $N_1$ decreased and was zero when the box was at $3.5\,\text{m}$; the common centre of mass was then at support 2, $x_\text{cm} = 3.0\,\text{m}$. Farther on, the plank toppled, because support 1 cannot pull the plank down. The values are calculated by the model: the plank is rigid and the supports are points.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Two equilibrium conditions
In the chapter Moment: the turning effect of a force we learned that a force rotates a body the more, the greater the moment of the force $M = Fr$. For a body to stay at rest, it is not enough that the forces cancel each other. A body is in equilibrium when it neither starts to move nor starts to rotate. For translation the condition is familiar from Newton's laws: the sum of the forces is zero. For rotation the condition is new: the sum of the moments of the forces is zero. In both cases in the figure two equal and opposite forces act on the ruler, so the sum of the forces is zero. On the left the forces lie along the same line and do not rotate the ruler. On the right they act along different lines, and their moments rotate the ruler in the same direction: the ruler starts to rotate, although its middle does not start to move.
Centre of mass
The Earth's gravitational pull acts on every part of a body. When the moment of the weight is calculated, however, the whole weight can be thought to act at a single point, the centre of mass of the body. The centre of mass of a uniform, symmetrical body, such as a plank of constant thickness, is at its middle. This is why the weight is drawn in a free-body diagram starting from the centre of mass. The common centre of mass of two bodies is between their centres of mass, nearer the heavier body. When a body is supported at one point, it stays in place only if the centre of mass is on the vertical line through the point of support: otherwise the moment of the weight about the point of support rotates the body.
When a body topples
A body standing on a surface rests on its whole base. It stays upright when the vertical line through the centre of mass falls within the base. The moment of the weight about the edge of the base then turns the body back onto its base. If the vertical line falls outside the base, the moment of the weight about the lower edge rotates the body over the edge, and the body topples. A low, wide body therefore withstands a greater tilt than a tall, narrow one. The boxes in the figure are on the same inclined plane, and friction prevents them from sliding.
Equilibrium conditions
A body is in equilibrium when both the sum of the forces and the sum of the moments of the forces are zero:
$$\begin{aligned} \Sigma\overline{F} &= \overline{0} \\ \Sigma M &= 0 \end{aligned}$$
where $\Sigma\overline{F}$ is the vector sum of the forces acting on the body ($\text{N}$) and $\Sigma M$ the sum of the moments of the forces about some point ($\text{Nm}$). The first is the equilibrium condition for translation and the second the moment condition, that is, the equilibrium condition for rotation. In components the first condition is $\Sigma F_x = 0$ and $\Sigma F_y = 0$. The moments are calculated as in the chapter Moment: the turning effect of a force: $M = Fr$, where $r$ is the lever arm of the force, and the sign tells the direction of rotation. In this chapter an anticlockwise moment is positive.
Choosing the reference point
A body at rest does not rotate about any point, so the moment condition holds about every point. It is worth choosing the point of application of an unknown force as the reference point for the moments: the lever arm of that force is then zero, so the force does not appear in the moment condition. If a force is not perpendicular to a rod-like body (a plank, a rod), it is resolved into components. When the reference point is on the line of the rod, the moment is given by the component perpendicular to the rod multiplied by the distance from the reference point. In the figure the force $\overline{F}$ acting on the end of the rod makes the angle $\alpha$ with the rod: the component $F\cos\alpha$ along the rod passes through the hinge and does not rotate the rod, and the perpendicular component $F\sin\alpha$ gives the moment $M = F\sin\alpha \cdot r$.
Two supports
The plank in the figure rests on two supports. Let us take the moments about the centre of mass: the lever arm of the weight $\overline{G}$ is zero, the support force $\overline{N}_1$ rotates the plank clockwise and $\overline{N}_2$ anticlockwise. From the moment condition $N_2 r_2 - N_1 r_1 = 0$ we get
$$N_1 r_1 = N_2 r_2$$
where $N_1$ and $N_2$ are the support forces ($\text{N}$) and $r_1$ and $r_2$ the distances of the supports from the centre of mass ($\text{m}$). The support nearer the centre of mass thus carries the larger share of the weight. From the condition for translation the sum of the support forces is $N_1 + N_2 = G$.
Common centre of mass
For two bodies of masses $m_1$ and $m_2$, the common centre of mass is at
$$x_\text{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$
where $x_1$ and $x_2$ are the positions of the centres of mass of the bodies ($\text{m}$) on the same axis. The formula follows from the moment condition: if the bodies are supported at $x_\text{cm}$, the moments of the weights about the point of support cancel each other. A loaded plank resting on two supports topples when the common centre of mass of the plank and the load moves outside the interval between the supports.
Common misconception
Many think that two supports always carry half of the weight each, or that the support farther away carries more because its "lever" is longer. The moment condition about the centre of mass shows the opposite: $N_1 r_1 = N_2 r_2$, so the support nearer the centre of mass carries more. Each support carries half of the weight only when the supports are at equal distances from the centre of mass.
Another mistake is to think that a loaded plank topples as soon as the load is past a support. The plank topples only when the common centre of mass of the plank and the load is past the support: until then the weight of the plank itself between the supports cancels the moment of the load.
A third mistake is to believe that moments must be calculated only about the centre of mass or a support. The moment condition holds about any point, and the same forces are obtained from different points. A good choice merely shortens the calculation.
Summary
- A body is in equilibrium when the sum of the forces is zero, $\Sigma\overline{F} = \overline{0}$ (for translation), and the sum of the moments of the forces is zero, $\Sigma M = 0$ (for rotation).
- The moment condition holds about any point. It is worth choosing the point of application of an unknown force as the reference point.
- The weight acts at the centre of mass, which is in the middle of a uniform, symmetrical body. For a plank on two supports $N_1 r_1 = N_2 r_2$: the support nearer the centre of mass carries more.
- A supported body stays upright when the vertical line through the centre of mass falls within the support area; otherwise it topples.
Optional Return to your predictions
Compare your predictions with the values of the model in the experiments and with the equilibrium conditions. Then return to the opening question: why can the painter push the can past the trestle onto the overhang, and what decides where the plank topples?
Would you change your prediction?
Apply
Worked example
Problem
Two carriers carry a uniform horizontal log of length $4.00\,\text{m}$ and mass $60.0\,\text{kg}$. Carrier A holds the front end of the log, and carrier B is $1.00\,\text{m}$ from the back end. With what force does each carrier hold the log? The carriers support the log vertically upwards.
Step 1/7
Given and find. The log: $L = 4.00\,\text{m}$, $m = 60.0\,\text{kg}$, $g = 9.81\,\text{m/s}^2$. The centre of mass is in the middle, $2.00\,\text{m}$ from A; B is $3.00\,\text{m}$ from A. We are asked for the forces of the carriers, $N_\text{A}$ and $N_\text{B}$.
Step 2/7
Principle. The log is at rest, so it is in equilibrium for both translation and rotation. The body is the log. In the free-body diagram the forces acting on it are the weight $\overline{G}$ downwards from the centre of mass and the forces of the carriers $\overline{N}_\text{A}$ and $\overline{N}_\text{B}$ upwards. A is chosen as the reference point, so the lever arm of $N_\text{A}$ is zero, and the anticlockwise moment is positive. The unknown $N_\text{B}$ is obtained from the moment condition and $N_\text{A}$ then from the sum of the forces.
Step 3/7
Formula solved. The moment condition about A and the condition $\Sigma F_y = 0$, with upwards positive:
$$\begin{aligned} &N_\text{B} r_\text{B} - G r_G = 0 \\ &N_\text{B} = \frac{m g r_G}{r_\text{B}} \\[6pt] &N_\text{A} + N_\text{B} - G = 0 \\ &N_\text{A} = m g - N_\text{B} \end{aligned}$$
Step 4/7
Unit conversions. Not needed: the mass is in kilograms and the lengths in metres. The lever arms of the forces are $r_G = 2.00\,\text{m}$ and $r_\text{B} = 3.00\,\text{m}$, because the forces are vertical and the log horizontal.
Step 5/7
Substitution. $$\begin{aligned} &G = m g \\ &= 60.0\,\text{kg} \cdot 9.81\,\text{m/s}^2 \\ &= 588.6\,\text{N} \\[6pt] &N_\text{B} = \frac{G r_G}{r_\text{B}} \\ &= \frac{588.6\,\text{N} \cdot 2.00\,\text{m}}{3.00\,\text{m}} \\ &= 392.4\,\text{N} \\[6pt] &N_\text{A} = G - N_\text{B} \\ &= 588.6\,\text{N} - 392.4\,\text{N} \\ &= 196.2\,\text{N} \end{aligned}$$
Step 6/7
Result. Carrier A holds the log with a force of $196\,\text{N}$ and carrier B with a force of $392\,\text{N}$. The results are given to three significant figures, like the mass $60.0\,\text{kg}$ and the lengths.
Step 7/7
Reasonableness. The sum of the forces is $196\,\text{N} + 392\,\text{N} = 588\,\text{N}$, the weight of the log. B is $1.00\,\text{m}$ and A $2.00\,\text{m}$ from the centre of mass, so B carries twice as much as A, as the moment condition about the centre of mass requires: $196\,\text{N} \cdot 2.00\,\text{m} = 392\,\text{N} \cdot 1.00\,\text{m}$.
Find the error in the free-body diagram
Draw the free-body diagram yourself
The picture shows a diving board at a swimming pool. The back end of the board is fastened to the edge of the pool with a bolt, and the board rests on a support beam $1.50\,\text{m}$ from the back end. A diver stands at the front end of the board. On paper, draw the free-body diagram of the board: the board as a rectangle and every force acting on it from its own point of application, labelled and roughly the right length. In which direction does the bolt pull the board?
Model diagram
The body is the board. Four forces act on it. The weight of the board $\overline{G}$ starts from the middle of the board, downwards. The diver presses the front end of the board down with the force $\overline{F}_\text{d}$; it is the force between the diver and the board, not the weight of the diver, which acts on the diver. The support beam pushes the board up with the force $\overline{N}$. The bolt pulls the back end of the board down with the force $\overline{F}_\text{bolt}$. Its direction can be seen from the moment condition about the support: $\overline{G}$ and $\overline{F}_\text{d}$ rotate the board clockwise, so the force at the back end must rotate it anticlockwise, and on the left of the support that is possible only with a force pointing downwards. $\overline{N}$ is the longest arrow in the picture, because the support force pointing upwards must cancel all three forces pointing downwards.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Optional Apply: the force of the bolt
The diving board of the previous task is uniform and horizontal. Its length is $4.00\,\text{m}$ and its mass $30.0\,\text{kg}$. The bolt is at the back end of the board and the support beam $1.50\,\text{m}$ from the back end. A diver of mass $60.0\,\text{kg}$ stands still at the front end of the board.
With what force does the bolt pull the board? Answer in newtons.
Hint 1: a guiding question
First draw the free-body diagram of the board. Which force drops out of the moment condition if the support beam is chosen as the reference point? What if the bolt is chosen?
Hint 2: what is the connection
The moment condition about the support beam: the lever arm of the force of the bolt is $1.50\,\text{m}$, that of the weight of the board $2.00\,\text{m} - 1.50\,\text{m} = 0.50\,\text{m}$ and that of the force of the diver $4.00\,\text{m} - 1.50\,\text{m} = 2.50\,\text{m}$.
Hint 3: an intermediate step
First calculate the moments on the right-hand side and divide their sum by the lever arm of the bolt, $1.50\,\text{m}$.
Show the solution
Given
board $L = 4.00\,\text{m}$, $m_\text{p} = 30.0\,\text{kg}$ (centre of mass $2.00\,\text{m}$ from the back end); support $1.50\,\text{m}$ from the back end; diver $m_\text{d} = 60.0\,\text{kg}$ at the front end; $g = 9.81\,\text{m/s}^2$
Find
the force of the bolt $F_\text{bolt}$ in newtons
Principle
The board is at rest, so it is in equilibrium for both translation and rotation, and the sum of the moments about any point is zero. The support is chosen as the reference point, so the unknown support force $N$ does not appear in the moment condition and the force of the bolt is obtained from a single equation.
Free-body diagram
The body is the board. The forces acting on it are the weight of the board $\overline{G}$ downwards from the middle (the Earth), the force of the diver $\overline{F}_\text{d}$ downwards from the front end (the diver is at rest, so $F_\text{d} = m_\text{d} g$), the force of the support $\overline{N}$ upwards and the force of the bolt $\overline{F}_\text{bolt}$ downwards from the back end. Anticlockwise moments are positive: the force of the bolt, on the left of the support, rotates anticlockwise; the weight of the board and the force of the diver, on the right of the support, rotate clockwise.
Formula
from which
Conversions
None needed, but the lever arms are measured from the support:
Substitution
The moments about the support are
together $1618.65\,\text{Nm}$. So
Answer
$1080\,\text{N}$
That is about $1.08\,\text{kN}$, three significant figures like the masses and lengths.
Reasonableness
A check from another reference point: about the bolt, the weight of the board gives the moment $294.3\,\text{N} \cdot 2.00\,\text{m} = 588.6\,\text{Nm}$ and the diver $588.6\,\text{N} \cdot 4.00\,\text{m} = 2354.4\,\text{Nm}$, together $2943\,\text{Nm}$, so
From the sum of the forces
the same force of the bolt. It is almost twice the weight of the diver, because the lever arm of the diver, $2.50\,\text{m}$, is more than one and a half times the lever arm of the bolt.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- Write the two equilibrium conditions of a body: for translation and for rotation.
Answer
For translation the sum of the forces is zero, $\Sigma\overline{F} = \overline{0}$. For rotation the sum of the moments of the forces is zero, $\Sigma M = 0$, about any point.
- The sum of the forces acting on a body is zero, but the sum of the moments is not. What happens to the body?
Answer
The body starts to rotate, although its centre of mass does not start to move. For example, a ruler with equal and opposite forces on its ends along different lines rotates on the spot.
1. Centre of mass and toppling
2. A seesaw
A seesaw is supported at its middle, so its own weight does not rotate it. A child of mass $24.0\,\text{kg}$ sits at one end of the board, $1.50\,\text{m}$ from the pivot. An adult keeps the board horizontal by pressing the other side down with a hand, vertically, $1.20\,\text{m}$ from the pivot.
With what force does the adult press? Answer in newtons.
Hint 1: a guiding question
Choose the pivot of the board as the reference point. Which forces then drop out of the moment condition?
Hint 2: what is the connection
The force of the child rotates the board in one direction and the force of the adult in the other: $F \cdot 1.20\,\text{m} = m g \cdot 1.50\,\text{m}$.
Show the solution
Given
child $m = 24.0\,\text{kg}$, lever arm $r_1 = 1.50\,\text{m}$; lever arm of the force of the adult $r_2 = 1.20\,\text{m}$; board supported at the middle; $g = 9.81\,\text{m/s}^2$
Find
the pressing force of the adult $F$ in newtons
Principle
The board stays horizontal at rest, so the sum of the moments about the pivot is zero. The pivot is a good reference point, because the lever arm of the support force and of the weight of the board is then zero.
Free-body diagram
The body is the board. The forces acting on it are the force of the child $\overline{F}_\text{c}$ downwards at one end (the child sits at rest, so the magnitude of the force is $F_\text{c} = mg$), the force of the adult $\overline{F}$ downwards on the other side, the weight of the board $\overline{G}$ downwards from the middle and the force of the support $\overline{N}$ upwards. $\overline{G}$ and $\overline{N}$ act at the pivot. The force of the child rotates the board in one direction and the force of the adult in the other; the direction of rotation of the force of the adult is chosen as positive.
Formula
$F r_2 - m g r_1 = 0$, from which $F = \dfrac{m g r_1}{r_2}$
Conversions
None needed: the mass is in kilograms and the lever arms in metres.
Substitution
Answer
$294\,\text{N}$
Three significant figures like the mass and the lever arms.
Reasonableness
The weight of the child is $235\,\text{N}$. The adult presses nearer the pivot, so the force is greater, $1.50 : 1.20 = 1.25$ times as large: $1.25 \cdot 235\,\text{N} \approx 294\,\text{N}$.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
Optional 3. Explain: why the nearer support carries more
A uniform plank rests on two supports, and one support is nearer the middle of the plank than the other. Using the equilibrium conditions, explain why the nearer support carries the larger share of the weight of the plank.
Model answer
The plank is at rest, so the sum of the forces is zero, $N_1 + N_2 = G$, and the sum of the moments about any point is zero. Take the moments about the centre of mass: the lever arm of the weight is zero, and the support forces rotate the plank in opposite directions. They cancel each other when $N_1 r_1 = N_2 r_2$, where $r_1$ and $r_2$ are the distances of the supports from the centre of mass. When $r_2 < r_1$, we must have $N_2 > N_1$: the force of the nearer support has the shorter lever arm, so it must be greater for its moment to be equal to that of the farther support.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Equilibrium
- A state in which neither the translational nor the rotational motion of an object changes.
- Moment
- A quantity that describes the turning effect of a force.$$M = Fr$$
- Centre of gravity
- The point at which the whole weight of an object can be thought to act.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 2-1Core
A crowbar is used as a lever to pull a nail. The pivot of the bar is $0.050\,\text{m}$ from the nail horizontally, and a hand presses the bar vertically down with a force of $80\,\text{N}$ at a horizontal distance of $0.60\,\text{m}$ from the pivot. The bar is light, so its weight can be ignored.
With what force does the bar pull the nail up when the bar is at rest?
Answer
$960\,\text{N}$
Show the solution
Given
force of the hand $F_\text{h} = 80\,\text{N}$, lever arm $r_\text{h} = 0.60\,\text{m}$ from the pivot; lever arm of the nail $r_\text{n} = 0.050\,\text{m}$ from the pivot; the bar is light and at rest
Find
the force $F_\text{n}$ with which the bar pulls the nail, in newtons
Principle
The bar stays at rest, so the sum of the moments about the pivot is zero. In the free-body diagram the forces acting on the bar are the force of the hand $F_\text{h}$ downwards, the force of the nail $F_\text{n}$ downwards on the other side of the pivot and the force of the support upwards at the pivot; the lever arm of the support force is zero, so it does not appear in the moment condition. The bar is light, so its weight is neglected. The force of the hand rotates the bar in one direction and the force of the nail in the other.
Formula
$F_\text{h} r_\text{h} - F_\text{n} r_\text{n} = 0$, from which
$F_\text{n}$$= \dfrac{F_\text{h} r_\text{h}}{r_\text{n}}$Conversions
None needed: the force is in newtons and the lever arms in metres.
Substitution
$F_\text{n}$$= \dfrac{80\,\text{N} \cdot 0.60\,\text{m}}{0.050\,\text{m}}$$= \dfrac{48\,\text{Nm}}{0.050\,\text{m}}$$= 960\,\text{N}$Answer
$960\,\text{N}$
Two significant figures like the force of the hand $80\,\text{N}$ and the lever arm $0.60\,\text{m}$.
Reasonableness
The lever arm of the nail is one twelfth of the lever arm of the hand, $0.60 : 0.050 = 12$, so the force pulling the nail is $12$ times the force of the hand: $12 \cdot 80\,\text{N} = 960\,\text{N}$. The lever multiplies the force, as a crowbar should.
- Exercise 2-2Core
A light ruler is supported at its middle. On its left, $0.30\,\text{m}$ from the pivot, hangs a mass whose weight is $0.50\,\text{N}$.
How large a vertical downward force is needed on the right, $0.20\,\text{m}$ from the pivot, to keep the ruler horizontal?
Answer
$0.75\,\text{N}$
Show the solution
Given
weight of the mass $F_1 = 0.50\,\text{N}$, lever arm $r_1 = 0.30\,\text{m}$; lever arm of the force asked $r_2 = 0.20\,\text{m}$; light ruler supported at the middle
Find
the force needed $F_2$ in newtons
Principle
The ruler stays at rest, so the sum of the moments about the pivot is zero. In the free-body diagram the forces acting on the ruler are the force of the mass $F_1$ downwards on the left, the force $F_2$ downwards on the right and the force of the support upwards at the pivot, whose lever arm is zero. The ruler is light, so its weight is neglected. $F_1$ rotates the ruler anticlockwise and $F_2$ clockwise.
Formula
$F_1 r_1 - F_2 r_2 = 0$, from which $F_2 = \dfrac{F_1 r_1}{r_2}$
Conversions
None needed: the force is in newtons and the lever arms in metres.
Substitution
$F_2$$= \dfrac{0.50\,\text{N} \cdot 0.30\,\text{m}}{0.20\,\text{m}}$$= 0.75\,\text{N}$Answer
$0.75\,\text{N}$
Two significant figures like the data.
Reasonableness
The force acts nearer the pivot than the mass, so it must be greater than the weight of the mass, $0.30 : 0.20 = 1.5$ times as large: $1.5 \cdot 0.50\,\text{N} = 0.75\,\text{N}$.
- Exercise 2-3
The combined weight of a wheelbarrow and its load is $620\,\text{N}$, and the common centre of mass is $0.400\,\text{m}$ horizontally from the axle of the wheel. The handles are $1.20\,\text{m}$ horizontally from the axle.
With what vertical force must the person pushing the wheelbarrow lift at the handles to keep the barrow at rest on the wheel and the handles?
Answer
$207\,\text{N}$
Show the solution
Given
weight of the wheelbarrow and load $G = 620\,\text{N}$, its lever arm from the axle $r_G = 0.400\,\text{m}$; lever arm of the handles $r_F = 1.20\,\text{m}$
Find
the lifting force $F$ in newtons
Principle
The wheelbarrow is at rest, so the sum of the moments about any point is zero. In the free-body diagram the forces acting on the wheelbarrow are the weight $G$ downwards from the centre of mass, the lifting force $F$ upwards at the handles and the support force of the ground upwards at the wheel. The axle of the wheel is chosen as the reference point, so the unknown support force drops out. The weight rotates the wheelbarrow clockwise and the lifting force anticlockwise.
Formula
$F r_F - G r_G = 0$, from which $F = \dfrac{G r_G}{r_F}$
Conversions
None needed: the weight is in newtons and the lever arms in metres.
Substitution
$F$$= \dfrac{620\,\text{N} \cdot 0.400\,\text{m}}{1.20\,\text{m}}$$= 206.7\,\text{N}$Answer
$207\,\text{N}$
Three significant figures like the lengths $0.400\,\text{m}$ and $1.20\,\text{m}$.
Reasonableness
The handles are three times as far from the axle as the centre of mass, so the lifting force is a third of the weight: $620\,\text{N} : 3 \approx 207\,\text{N}$. The remaining $413\,\text{N}$ is carried by the wheel.
- Exercise 2-4
A uniform plank of mass $12.0\,\text{kg}$ rests horizontally on two supports. The supports are at the ends of the plank.
How large is the support force of each support?
Answer
$58.9\,\text{N}$
Show the solution
Given
uniform plank $m = 12.0\,\text{kg}$; supports at the ends of the plank; $g = 9.81\,\text{m/s}^2$
Find
the support force $N$ of each support in newtons
Principle
The plank is at rest, so the sum of the forces and the sum of the moments are zero. In the free-body diagram the forces acting on the plank are the weight $G$ downwards from the middle and the support forces $N_1$ and $N_2$ upwards at the ends. The centre of mass is in the middle, so the supports are at equal distances from it, and the moment condition about the centre of mass, $N_1 r = N_2 r$, gives $N_1 = N_2 = N$. The support forces are then obtained from the sum of the forces.
Formula
$N_1 + N_2 - G = 0$ and $N_1 = N_2 = N$, so $N = \dfrac{m g}{2}$
Conversions
None needed: the mass is in kilograms.
Substitution
$N$$= \dfrac{12.0\,\text{kg} \cdot 9.81\,\text{m/s}^2}{2}$$= \dfrac{117.72\,\text{N}}{2}$$= 58.86\,\text{N}$Answer
$58.9\,\text{N}$
Three significant figures like the mass $12.0\,\text{kg}$.
Reasonableness
The sum of the support forces $2 \cdot 58.86\,\text{N} = 117.7\,\text{N}$ is the weight of the plank. The weight of a plank of about $12\,\text{kg}$ is about $120\,\text{N}$, so each support carries about $60\,\text{N}$.
- Exercise 2-5
At one end of a light rod of length $0.80\,\text{m}$ there is a mass of $2.0\,\text{kg}$ and at the other end a mass of $6.0\,\text{kg}$.
How far from the lighter mass is the common centre of mass of the rod, that is, at which point must the rod be supported to stay horizontal?
Answer
$0.60\,\text{m}$
Show the solution
Given
light rod $L = 0.80\,\text{m}$; masses $m_1 = 2.0\,\text{kg}$ at $x_1 = 0$ and $m_2 = 6.0\,\text{kg}$ at $x_2 = 0.80\,\text{m}$
Find
the position of the common centre of mass $x_\text{cm}$ measured from the lighter mass
Principle
The rod stays horizontal when it is supported at the common centre of mass: the moments of the weights of the masses about the point of support then cancel each other. In the free-body diagram the forces acting on the rod are the forces of the masses $m_1 g$ and $m_2 g$ downwards at the ends and the force of the support upwards at the point of support. The rod is light, so its weight is neglected.
Formula
$x_\text{cm}$$= \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$Conversions
None needed: the unit of mass cancels, and the length is in metres.
Substitution
$x_\text{cm}$$= \dfrac{2.0\,\text{kg} \cdot 0\,\text{m} + 6.0\,\text{kg} \cdot 0.80\,\text{m}}{2.0\,\text{kg} + 6.0\,\text{kg}}$$= \dfrac{4.8\,\text{kg}\cdot\text{m}}{8.0\,\text{kg}}$$= 0.60\,\text{m}$Answer
$0.60\,\text{m}$
From the lighter mass, two significant figures like the masses and the length.
Reasonableness
The centre of mass is nearer the heavier mass, $0.20\,\text{m}$ from it. The moments about the point of support cancel:
$2.0\,\text{kg} \cdot 0.60\,\text{m}$$= 6.0\,\text{kg} \cdot 0.20\,\text{m}$$= 1.2\,\text{kg}\cdot\text{m}$
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 2-6Core
A uniform plank of length $4.00\,\text{m}$ and mass $25.0\,\text{kg}$ rests horizontally on two supports. The supports are $0.50\,\text{m}$ and $3.00\,\text{m}$ from the left end of the plank.
How large is the support force of the left support?
Answer
$98.1\,\text{N}$
Show the solution
Given
plank $L = 4.00\,\text{m}$, $m = 25.0\,\text{kg}$, centre of mass at $2.00\,\text{m}$; supports at $x_1 = 0.50\,\text{m}$ and $x_2 = 3.00\,\text{m}$; $g = 9.81\,\text{m/s}^2$
Find
the support force of the left support $N_1$ in newtons
Principle
The plank is at rest, so the sum of the moments about any point is zero. In the free-body diagram the forces acting on the plank are the weight $G$ downwards from the middle and the support forces $N_1$ and $N_2$ upwards. The right support is chosen as the reference point, so the unknown $N_2$ drops out of the moment condition. The weight is to the left of the right support and rotates the plank anticlockwise, $N_1$ clockwise.
Formula
$G r_G - N_1 r_1 = 0$, from which $N_1 = \dfrac{m g\, r_G}{r_1}$
Conversions
None needed, but the lever arms are calculated from the right support:
$r_G$$= 3.00\,\text{m} - 2.00\,\text{m}$$= 1.00\,\text{m}$$r_1$$= 3.00\,\text{m} - 0.50\,\text{m}$$= 2.50\,\text{m}$Substitution
$N_1$$= \dfrac{25.0\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.00\,\text{m}}{2.50\,\text{m}}$$= \dfrac{245.25\,\text{Nm}}{2.50\,\text{m}}$$= 98.1\,\text{N}$Answer
$98.1\,\text{N}$
Three significant figures like the data.
Reasonableness
From the sum of the forces
$N_2$$= 245.25\,\text{N} - 98.1\,\text{N}$$= 147.15\,\text{N}$About the centre of mass $N_1 \cdot 1.50\,\text{m} = 147.2\,\text{Nm}$ and $N_2 \cdot 1.00\,\text{m} = 147.2\,\text{Nm}$: the moments cancel, and the farther support carries less, as it should.
- Exercise 2-7Core
A uniform plank of length $3.0\,\text{m}$ and mass $20\,\text{kg}$ has been laid across a ditch, with both ends resting on the ground. A person of mass $70\,\text{kg}$ stands on the plank $1.0\,\text{m}$ from the left end.
How large is the support force at the left end of the plank?
Answer
$560\,\text{N}$
Show the solution
Given
plank $L = 3.0\,\text{m}$, $m_\text{p} = 20\,\text{kg}$, centre of mass in the middle; person $m_\text{pers} = 70\,\text{kg}$ at $1.0\,\text{m}$ from the left end; supports at the ends; $g = 9.81\,\text{m/s}^2$
Find
the support force of the left end $N_1$ in newtons
Principle
The plank is at rest, so the sum of the moments about any point is zero. In the free-body diagram the forces acting on the plank are its weight $G_\text{p}$ downwards from the middle, the force of the person $F_\text{pers} = m_\text{pers} g$ downwards and the support forces $N_1$ and $N_2$ upwards at the ends. The right end is chosen as the reference point, so $N_2$ drops out. $G_\text{p}$ and $F_\text{pers}$ rotate the plank about it anticlockwise, $N_1$ clockwise.
Formula
$m_\text{p} g\, r_\text{p} + m_\text{pers} g\, r_\text{pers} - N_1 L = 0$, from which
$N_1$$= \dfrac{(m_\text{p} r_\text{p} + m_\text{pers} r_\text{pers})\, g}{L}$Conversions
None needed, but the lever arms are calculated from the right end:
$r_\text{p}$$= 1.5\,\text{m}$$r_\text{pers}$$= 3.0\,\text{m} - 1.0\,\text{m}$$= 2.0\,\text{m}$Substitution
The moments are
$m_\text{p} g\, r_\text{p}$$= 20\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.5\,\text{m}$$= 294.3\,\text{Nm}$$m_\text{pers} g\, r_\text{pers}$$= 70\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 2.0\,\text{m}$$= 1373.4\,\text{Nm}$together $1667.7\,\text{Nm}$. So
$N_1$$= \dfrac{1667.7\,\text{Nm}}{3.0\,\text{m}}$$= 555.9\,\text{N}$Answer
$560\,\text{N}$
Two significant figures like the data.
Reasonableness
The total weight is $90\,\text{kg} \cdot 9.81\,\text{m/s}^2 = 882.9\,\text{N}$, so the right end carries $327\,\text{N}$. The left end carries more, because the person stands nearer to it.
- Exercise 2-8
A uniform bookshelf has depth $0.400\,\text{m}$ and height $1.80\,\text{m}$, and its centre of mass is in the middle. The shelf is tilted forwards so that it rests only on its front edge.
What is the angle of tilt from the vertical when the shelf is at the limit of toppling, that is, when it would topple by itself if the angle grew any more? Answer in degrees.
Answer
$12.5^\circ$
Show the solution
Given
depth of the shelf $d = 0.400\,\text{m}$, height $h = 1.80\,\text{m}$, centre of mass in the middle
Find
the angle of tilt $\alpha$ from the vertical at the limit of toppling, in degrees
Principle
In the figure the angle $\alpha$ is the angle between the vertical and the height direction of the shelf, that is, its front edge, so it is the tilt of the shelf from the vertical. A shelf resting on its front edge stays in place when the vertical line through the centre of mass falls behind the edge: the moment of the weight about the edge then turns the shelf back. The limit is when the centre of mass is exactly above the edge, and the moment of the weight about the edge is zero. In the free-body diagram the only forces acting on the shelf are the weight downwards from the centre of mass and the support force of the floor upwards at the edge.
Formula
At the limit the segment from the centre of mass to the front edge $P$ is vertical. The centre of mass is in the middle of the shelf, so it is half the height, $\tfrac{h}{2}$, from the middle of the base along the height direction of the shelf, and the middle of the base is half the depth, $\tfrac{d}{2}$, from the edge $P$. This gives a right-angled triangle whose legs are $\tfrac{d}{2}$ and $\tfrac{h}{2}$ and whose hypotenuse is the vertical segment. At the centre of mass the angle between the vertical and the leg $\tfrac{h}{2}$ is $\alpha$, because the leg $\tfrac{h}{2}$ is along the front edge of the shelf, and the leg opposite the angle is $\tfrac{d}{2}$. So
$\tan\alpha$$= \dfrac{d/2}{h/2}$$= \dfrac{d}{h}$from which $\alpha = \arctan\dfrac{d}{h}$
Conversions
None needed: both lengths are in metres, and the unit cancels.
Substitution
$\alpha$$= \arctan\dfrac{0.400\,\text{m}}{1.80\,\text{m}}$$= \arctan 0.2222$$= 12.53^\circ$Answer
$12.5^\circ$
Three significant figures like the dimensions $0.400\,\text{m}$ and $1.80\,\text{m}$.
Reasonableness
A tall, narrow shelf topples already at a small tilt. If the shelf were as deep as it is tall, the limit would be $45^\circ$; the depth is now less than a quarter of the height, so the angle must be much smaller.
- Exercise 2-9
A forearm is horizontal, and the hand holds a $5.0\,\text{kg}$ kettlebell $0.32\,\text{m}$ from the elbow joint. The weight of the forearm and hand is $15\,\text{N}$, and its centre of mass is $0.15\,\text{m}$ from the elbow joint. The biceps pulls the forearm vertically up at a distance of $0.040\,\text{m}$ from the elbow joint.
How large is the force of the biceps?
Answer
$450\,\text{N}$
Show the solution
Given
kettlebell $m = 5.0\,\text{kg}$, lever arm $r_\text{kb} = 0.32\,\text{m}$; weight of the forearm $G_\text{f} = 15\,\text{N}$, lever arm $r_\text{f} = 0.15\,\text{m}$; lever arm of the muscle $r_\text{m} = 0.040\,\text{m}$; $g = 9.81\,\text{m/s}^2$
Find
the force of the biceps $F$ in newtons
Principle
The forearm is at rest, so the sum of the moments about the elbow joint is zero. In the free-body diagram the forces acting on the forearm are the force of the kettlebell $mg$ downwards from the palm, the weight of the forearm downwards from its centre of mass, the force of the muscle $F$ upwards and the force of the elbow joint, which points down, because the force of the muscle is greater than the weights together. The joint is chosen as the reference point, so the unknown force of the joint drops out. The muscle rotates the forearm in one direction, the weights in the other.
Formula
$F r_\text{m} - m g\, r_\text{kb} - G_\text{f} r_\text{f} = 0$, from which
$F$$= \dfrac{m g\, r_\text{kb} + G_\text{f} r_\text{f}}{r_\text{m}}$Conversions
None needed: the lever arms are in metres and the mass in kilograms.
Substitution
The moments are
$m g\, r_\text{kb}$$= 5.0\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 0.32\,\text{m}$$= 15.696\,\text{Nm}$$G_\text{f} r_\text{f}$$= 15\,\text{N} \cdot 0.15\,\text{m}$$= 2.25\,\text{Nm}$together $17.946\,\text{Nm}$. So
$F$$= \dfrac{17.946\,\text{Nm}}{0.040\,\text{m}}$$= 448.65\,\text{N}$Answer
$450\,\text{N}$
Two significant figures like the data.
Reasonableness
The lever arm of the muscle is only an eighth of the lever arm of the kettlebell, so the force of the muscle is many times the weight of the kettlebell, $49\,\text{N}$: the kettlebell alone requires $8 \cdot 49\,\text{N} \approx 390\,\text{N}$, and the weight of the forearm adds about $56\,\text{N}$. From the sum of the forces the force of the joint is downwards, $448.65\,\text{N} - 49.05\,\text{N} - 15\,\text{N} = 384.6\,\text{N}$: the joint is compressed, as it should be when the muscle pulls the forearm up.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 2-10
A painter lays a uniform plank ($4.00\,\text{m}$, $20.0\,\text{kg}$) across two trestles so that the trestles are $0.50\,\text{m}$ and $3.00\,\text{m}$ from the left end of the plank. The mass of the painter is $60.0\,\text{kg}$.
How far from the left end of the plank can the painter walk, past the right-hand trestle, before the plank starts to topple?
Answer
$3.33\,\text{m}$
Show the solution
Given
plank $L = 4.00\,\text{m}$, $m_\text{p} = 20.0\,\text{kg}$, centre of mass $x_\text{p} = 2.00\,\text{m}$; trestles $x_1 = 0.50\,\text{m}$ and $x_2 = 3.00\,\text{m}$ from the left end; painter $m_\text{pt} = 60.0\,\text{kg}$
Find
the greatest distance $x$ of the painter from the left end of the plank before toppling
Principle
The plank stays on the trestles until the support force $N_1$ of the left trestle drops to zero, because a trestle can only push upwards. In the free-body diagram the forces acting on the plank are the weight $G_\text{p}$ downwards from the middle, the force of the painter $m_\text{pt} g$ downwards and the support forces upwards. The right trestle is chosen as the reference point: at the limit $N_1 = 0$, and the moment of the weight of the plank (anticlockwise) cancels the moment of the painter (clockwise).
Formula
$m_\text{p} g\,(x_2 - x_\text{p}) - m_\text{pt} g\,(x - x_2) = 0$, from which
$x$$= x_2 + \dfrac{m_\text{p}\,(x_2 - x_\text{p})}{m_\text{pt}}$Conversions
None needed: $g$ and the unit of mass cancel, and the lengths are in metres. The lever arm of the weight of the plank from the trestle is
$x_2 - x_\text{p}$$= 3.00\,\text{m} - 2.00\,\text{m}$$= 1.00\,\text{m}$Substitution
$x$$= 3.00\,\text{m} + \dfrac{20.0\,\text{kg} \cdot 1.00\,\text{m}}{60.0\,\text{kg}}$$= 3.00\,\text{m} + 0.333\,\text{m}$$= 3.333\,\text{m}$Answer
$3.33\,\text{m}$
From the left end of the plank, three significant figures like the data.
Reasonableness
At the limit the common centre of mass of the plank and the painter is at the trestle: the sum of the products of mass and position is
$20.0\,\text{kg} \cdot 2.00\,\text{m} + 60.0\,\text{kg} \cdot 3.333\,\text{m}$$= 240\,\text{kg}\cdot\text{m}$and divided by the total mass $80.0\,\text{kg}$ it gives $3.00\,\text{m}$. The painter can walk only $0.33\,\text{m}$ past the trestle, because he is three times as heavy as the plank.
- Exercise 2-11
A shop sign hangs from the end of a horizontal uniform rod. The rod has length $1.20\,\text{m}$ and mass $4.00\,\text{kg}$, and its other end is attached to a wall with a hinge. The sign, of mass $8.00\,\text{kg}$, hangs from the outer end of the rod. A cable runs from the outer end of the rod to the wall above the hinge so that the angle between the cable and the rod is $30.0^\circ$.
How large is the tension force of the cable?
Answer
$196\,\text{N}$
Show the solution
Given
rod $L = 1.20\,\text{m}$, $m_\text{r} = 4.00\,\text{kg}$, centre of mass in the middle; sign $m_\text{s} = 8.00\,\text{kg}$ at the outer end; cable attached to the wall above the hinge, angle between the cable and the rod $\alpha = 30.0^\circ$; $g = 9.81\,\text{m/s}^2$
Find
the tension force of the cable $T$ in newtons
Principle
The rod is at rest, so the sum of the moments about the hinge is zero. In the free-body diagram the forces acting on the rod are its weight downwards from the middle, the force of the sign $m_\text{s} g$ downwards from the outer end, the tension force of the cable $\overline{T}$ from the outer end along the cable towards the wall, and the force of the hinge. The hinge is chosen as the reference point, so its unknown force drops out. $\overline{T}$ is resolved into components: the component along the rod passes through the hinge and causes no moment, and the perpendicular component $T \sin\alpha$ rotates the rod upwards.
Formula
$T \sin\alpha \cdot L - m_\text{r} g \cdot \tfrac{L}{2} - m_\text{s} g \cdot L$$= 0$from which
$T$$= \dfrac{(\tfrac{1}{2} m_\text{r} + m_\text{s})\, g}{\sin\alpha}$Conversions
None needed: the masses are in kilograms, and the length $L$ cancels.
Substitution
$T$$= \dfrac{(2.00\,\text{kg} + 8.00\,\text{kg}) \cdot 9.81\,\text{m/s}^2}{\sin 30.0^\circ}$$= \dfrac{98.1\,\text{N}}{0.500}$$= 196.2\,\text{N}$Answer
$196\,\text{N}$
Three significant figures like the data.
Reasonableness
The vertical component of the cable force, $T \sin 30.0^\circ = 98.1\,\text{N}$, is less than the combined weight of the rod and the sign, $117.7\,\text{N}$; the hinge carries the rest. A shallow cable requires a large tension: as the angle approaches zero, $T$ would grow without limit.
- Exercise 2-12
Two people carry a uniform log of length $5.00\,\text{m}$ and mass $90.0\,\text{kg}$ horizontally. The front carrier holds the front end of the log, and can lift with a force of at most $300\,\text{N}$.
How far from the back end of the log must the rear carrier hold the log so that the force of the front carrier is exactly $300\,\text{N}$?
Answer
$1.21\,\text{m}$
Show the solution
Given
log $L = 5.00\,\text{m}$, $m = 90.0\,\text{kg}$, centre of mass in the middle; force of the front carrier $F = 300\,\text{N}$ at the front end; $g = 9.81\,\text{m/s}^2$
Find
the distance $a$ of the rear carrier from the back end of the log
Principle
The log is at rest, so the sum of the moments about any point is zero. In the free-body diagram the forces acting on the log are the weight $G$ downwards from the middle and the forces of the carriers upwards. The rear carrier is chosen as the reference point, so his unknown force drops out. The centre of mass is between the carriers, so the weight and the force of the front carrier rotate the log in opposite directions.
Formula
$F\,(L - a) - m g\,(\tfrac{L}{2} - a) = 0$, from which
$a$$= \dfrac{m g \cdot \tfrac{L}{2} - F L}{m g - F}$Conversions
None needed: the mass is in kilograms and the length in metres. The weight is
$G$$= 90.0\,\text{kg} \cdot 9.81\,\text{m/s}^2$$= 882.9\,\text{N}$Substitution
The numerator is
$882.9\,\text{N} \cdot 2.50\,\text{m} - 300\,\text{N} \cdot 5.00\,\text{m}$$= 707.25\,\text{Nm}$and the denominator $882.9\,\text{N} - 300\,\text{N} = 582.9\,\text{N}$. So
$a$$= \dfrac{707.25\,\text{Nm}}{582.9\,\text{N}}$$= 1.213\,\text{m}$Answer
$1.21\,\text{m}$
From the back end of the log, three significant figures like the data.
Reasonableness
The rear carrier carries $882.9\,\text{N} - 300\,\text{N} = 582.9\,\text{N}$, and about the centre of mass the moments cancel: $300\,\text{N} \cdot 2.50\,\text{m} = 750\,\text{Nm}$ and $582.9\,\text{N} \cdot (2.50 - 1.213)\,\text{m} = 750\,\text{Nm}$. The rear carrier is nearer the centre of mass and carries more.