Back to homePhysics and Chemistry · 1st Bach. Science · Change subject
Start for free
BAC1 tech · Physics and Chemistry
Statics and dynamics · Statics

The rigid body and the moment of a force

A door handle far from the hinges opens the door with little effort; a long spanner loosens a nut that your bare hand cannot budge; two equal and opposite forces on a steering wheel make it turn even though their vector sum is zero. To explain all this we need a new quantity, one that did not appear when we dealt with a point particle: the moment of a force.

Before you start, make sure you know: Equilibrium of forces (2nd ESO)
Before you start, make sure you know: The vector nature of forces (4th ESO)

The moment of a force: M = F · d

Push a door with one finger close to the hinges and then do the same next to the handle: with the same force, the further from the hinges you push, the more easily it opens. How do we measure the ability of a force to cause a rotation, depending on where it is applied?

Moment of a perpendicular force applied at a distance d from an axis of rotation. axis d F M = F · d
Practice★☆☆Not started
To open a door, you push perpendicularly to it with a force of $25\ \text{N}$ applied at $0,8\ \text{m}$ from the hinges, as shown in the figure (top view). Calculate…

To open a door, you push perpendicularly to it with a force of $25\ \text{N}$ applied at $0,8\ \text{m}$ from the hinges, as shown in the figure (top view).

door hinges F = 25 N 0,8 m M = ?

Calculate the moment of the force about the axis of the hinges, $M = F \cdot d$, and state its unit in the International System of Units.

Solve
Practice★☆☆Not started
A child pushes a merry-go-round by directing his force straight towards the central axis, and no matter how hard he pushes, he cannot get it to rotate. The situation is…

A child pushes a merry-go-round by directing his force straight towards the central axis, and no matter how hard he pushes, he cannot get it to rotate. The situation is shown in the figure (top view).

merry-go-round axis F M = ?
  1. a) Explain, using the definition of the moment of a force, why the moment he produces is zero.
  2. b) State how he should direct his push to obtain maximum moment with the same force.
Solve
Practice★★☆Not started
The wind pushes a portable solar panel with a force of $50\ \text{N}$ applied at $0,9\ \text{m}$ from its pivot axis and making an angle of $30^\circ$ with the arm of…

The wind pushes a portable solar panel with a force of $50\ \text{N}$ applied at $0,9\ \text{m}$ from its pivot axis and making an angle of $30^\circ$ with the arm of the panel, as shown in the figure.

axis panel F = 50 N 30° 0,9 m M = ?

When the force is not perpendicular to the arm, the moment is given by $M = F \cdot d \cdot \sen \alpha$ (use $\sen 30^\circ = 0,5$).

  1. a) Calculate the moment of the wind force about the axis.
  2. b) Reason for which angle of incidence the moment would be maximum with the same force.
Solve

Sign of the moment and rotational equilibrium

A moment makes a body turn clockwise or anticlockwise. To add moments together, we give each direction a sign, just as we do with forces along a line.

A central axis with two moments in opposite directions: M₁ anticlockwise (positive) and M₂ clockwise (negative). axis O M₁ anticlockwise (+) M₂ clockwise (−)
Practice★☆☆Not started
Two children play with a hotel revolving door. One pushes it clockwise producing a moment of $12\ \text{N}\cdot\text{m}$ and the other pushes it anticlockwise producing…

Two children play with a hotel revolving door. One pushes it clockwise producing a moment of $12\ \text{N}\cdot\text{m}$ and the other pushes it anticlockwise producing a moment of $12\ \text{N}\cdot\text{m}$.

  1. a) Taking the anticlockwise direction as positive, assign a sign to each moment.
  2. b) Calculate the total moment on the door and state whether it will start to rotate (the door was initially at rest).
Solve
Practice★★☆Not started
Two workers rotate the handwheel of an industrial valve by pushing in opposite directions: one exerts $40\ \text{N}$ at $0,5\ \text{m}$ from the axis anticlockwise and…

Two workers rotate the handwheel of an industrial valve by pushing in opposite directions: one exerts $40\ \text{N}$ at $0,5\ \text{m}$ from the axis anticlockwise and the other exerts $50\ \text{N}$ at $0,3\ \text{m}$ from the axis clockwise.

  1. a) Calculate the moment of each force, assigning a positive sign to the anticlockwise direction.
  2. b) Calculate the resultant moment on the handwheel and state which direction it will rotate in.
Solve
Practice★★★Not started
To open the gate of an industrial warehouse, a moment of $24\ \text{N}\cdot\text{m}$ about its hinges is required, and the handle is at $0,8\ \text{m}$ from the axis.…

To open the gate of an industrial warehouse, a moment of $24\ \text{N}\cdot\text{m}$ about its hinges is required, and the handle is at $0,8\ \text{m}$ from the axis. Use $\sen 30^\circ = 0,5$.

  1. a) Calculate the minimum force required if pulling perpendicularly to the gate.
  2. b) Calculate the required force if pulling at an angle of $30^\circ$ to the gate (the moment is then $M = F \cdot d \cdot \sen \alpha$).
  3. c) Reason what would happen if pulling parallel to the gate and conclude what factor, in addition to the magnitude of the force and the distance, determines the moment.
Solve

Complete equilibrium of a rigid body

If two people push a table with opposite forces along the same line, it does not move; but if they push at diagonally opposite corners, the table starts to turn even though the sum of the forces is still zero. What conditions must be met at the same time for a body neither to move nor to start rotating?

Steering wheel with two equal, parallel forces in opposite directions, tangential to the ends of a diameter: the net force is zero but the net moment makes the wheel turn. F F axis ΣF = 0, but ΣM ≠ 0: the wheel turns
Practice★☆☆Not started
Two equal and opposite forces act on a car's steering wheel, one at each end of a diameter: the sum of the forces is zero, yet the steering wheel turns. a) Explain why…

Two equal and opposite forces act on a car's steering wheel, one at each end of a diameter: the sum of the forces is zero, yet the steering wheel turns.

  1. a) Explain why the condition $\sum \vec{F} = 0$ is not sufficient to guarantee the equilibrium of a rigid body.
  2. b) State the two conditions that a rigid body must satisfy to be in complete equilibrium.
Solve
Practice★★★Not started
A uniform walkway of $5\ \text{m}$ and $80\ \text{kg}$ is horizontally supported on two supports at its ends. A person of $60\ \text{kg}$ is standing on the walkway, at…

A uniform walkway of $5\ \text{m}$ and $80\ \text{kg}$ is horizontally supported on two supports at its ends. A person of $60\ \text{kg}$ is standing on the walkway, at $1\ \text{m}$ from the left-hand end, as shown in the figure.

walkway: 80 kg 60 kg R₁ = ? R₂ = ? 1 m 5 m

Take $g \approx 10\ \text{m/s}^2$.

  1. a) Calculate the weight of the footbridge and that of the person, and state at which point each acts.
  2. b) Applying the condition of moment equilibrium about the left support, calculate the force exerted by the right support.
  3. c) Applying $\sum \vec{F} = 0$, calculate the force exerted by the left support.
  4. d) Check the result by verifying the condition of moments about the right support.
Solve

The lever

To lift a heavy rock, it is enough to wedge a rigid bar over a small stone and push the long end with a moderate effort. What mathematical law relates the forces and their distances from the pivot so that our force is multiplied?

First-class lever in equilibrium: a rigid bar resting on the fulcrum, with the effort downwards on the left, the load downwards on the right and the moment arms measured from the fulcrum. effort load fulcrum d effort d load F effort · d effort = F load · d load
Practice★☆☆Not started
In a garden wheelbarrow loaded with soil: a) Identify where the fulcrum, effort and load are located. b) State which element lies between the other two and deduce which…

In a garden wheelbarrow loaded with soil:

  1. a) Identify where the fulcrum, effort and load are located.
  2. b) State which element lies between the other two and deduce which class of lever it is.
Solve
Practice★★☆Not started
A builder's wheelbarrow carries a load of $480\ \text{N}$ located at $0,5\ \text{m}$ from the wheel axle, and the operator holds the handles at $1,5\ \text{m}$ from the…

A builder's wheelbarrow carries a load of $480\ \text{N}$ located at $0,5\ \text{m}$ from the wheel axle, and the operator holds the handles at $1,5\ \text{m}$ from the axle, as shown in the figure.

axle 480 N F = ? 0,5 m 1,5 m
  1. a) Calculate the vertical force that the operator must exert to maintain the wheelbarrow in equilibrium.
  2. b) Compare this force with the weight of the load and explain what mechanical advantage this lever offers.
Solve
Practice★★☆Not started
A nutcracker exerts a load force of $150\ \text{N}$ on a nut, applied at $0,03\ \text{m}$ from the fulcrum, and the hand squeezes at $0,12\ \text{m}$ from the fulcrum.…

A nutcracker exerts a load force of $150\ \text{N}$ on a nut, applied at $0,03\ \text{m}$ from the fulcrum, and the hand squeezes at $0,12\ \text{m}$ from the fulcrum.

  1. a) Calculate the moment exerted by the nut about the fulcrum, in $\text{N}\cdot\text{m}$.
  2. b) Calculate the minimum force the hand must exert to crack the nut.
Solve
Practice★★★Not started
A person holds a dumbbell of $4\ \text{kg}$ with their forearm horizontal. The dumbbell is at $32\ \text{cm}$ from the elbow joint and the biceps pulls the forearm at…

A person holds a dumbbell of $4\ \text{kg}$ with their forearm horizontal. The dumbbell is at $32\ \text{cm}$ from the elbow joint and the biceps pulls the forearm at $4\ \text{cm}$ from the elbow, as shown in the diagram.

elbow F = ? 4 kg P 4 cm 32 cm

Take $g \approx 10\ \text{m/s}^2$.

  1. a) Calculate the moment of the weight of the dumbbell about the elbow.
  2. b) Calculate the force that the biceps must exert to maintain equilibrium.
  3. c) Compare that force with the weight of the dumbbell, explain why the muscle exerts much more force than the supported weight, and what is gained in return with this lever.
Solve

The centre of gravity

Uniform ruler resting on a support: the centre of gravity is at its geometric centre and the weight P acts vertically downwards, falling within the base of support. CG P base of support
Practice★☆☆Not started
a) Where is the centre of gravity of a uniform ruler approximately located? Describe how you could verify this by balancing it on a finger. b) Explain what happens to an…
  1. a) Where is the centre of gravity of a uniform ruler approximately located? Describe how you could verify this by balancing it on a finger.
  2. b) Explain what happens to an object resting on the ground when the vertical line through its centre of gravity falls outside its base of support, and give an everyday example.
Solve
Practice★☆☆Not started
On a see-saw, a girl of $40\ \text{kg}$ sits at $1,5\ \text{m}$ from the pivot axis and another girl of $30\ \text{kg}$ sits at $2\ \text{m}$ from the pivot, on the…

On a see-saw, a girl of $40\ \text{kg}$ sits at $1,5\ \text{m}$ from the pivot axis and another girl of $30\ \text{kg}$ sits at $2\ \text{m}$ from the pivot, on the opposite side, as shown in the figure.

40 kg 30 kg 1,5 m 2 m ΣM = ?

Take $g \approx 10\ \text{m/s}^2$.

  1. a) Calculate the moment of each girl's weight about the pivot, stating the direction of rotation each produces.
  2. b) Verify that the sum of moments is zero and deduce whether the see-saw is in equilibrium.
Solve
Practice★★☆Not started
A uniform bar of $3\ \text{m}$ is used as a see-saw: a person of $60\ \text{kg}$ sits at the left-hand end and a person of $20\ \text{kg}$ sits at the right-hand end, as…

A uniform bar of $3\ \text{m}$ is used as a see-saw: a person of $60\ \text{kg}$ sits at the left-hand end and a person of $20\ \text{kg}$ sits at the right-hand end, as shown in the figure.

60 kg 20 kg x = ? 3 m

Take $g \approx 10\ \text{m/s}^2$ and ignore the weight of the bar.

  1. a) Set up the equilibrium condition of moments about the pivot point, letting $x$ be the distance from this point to the left-hand end.
  2. b) Solve the equation and calculate where the pivot must be placed.
  3. c) Explain why the pivot point ends up closer to the heavier person.
Solve
Practice★★★Not started
a) Explain in your own words what the centre of gravity of a body is. b) A delivery van travels with several heavy packages tied to the roof rack. Explain, using the…
  1. a) Explain in your own words what the centre of gravity of a body is.
  2. b) A delivery van travels with several heavy packages tied to the roof rack. Explain, using the moment of the weight and the position of the centre of gravity, why it is less stable when taking bends or driving on side slopes than if the load were on the floor of the cabin.
  3. c) In a rugby maul, players bend their legs and spread their feet apart. Interpret this posture in terms of centre of gravity and base of support.
Solve

Back to the beginning: with the spanner you increase the distance from the axis ($d$), so that the same force from your hand produces a much greater moment ($M = F \cdot d$) than with your bare fingers, overcoming the resistance of the nut.

Summary

See all the exercises in this section