Simple machines and torque

A simple machine does not create work: it trades force for displacement. A lever does this by rotating about a pivot (fulcrum), and the rotational effect of a force is described by its torque: $\tau = rF\sin\theta,$ where $r$ is the distance from the pivot to the point of application (the position vector) and $\theta$ is the angle between the force and the line from the pivot to the point of application. A lever is in rotational equilibrium when the torques of the effort and the load cancel: $F_1 l_1 = F_2 l_2,$ where $l_1$ and $l_2$ are the perpendicular distances from the fulcrum (the lever arms). This gives the ideal mechanical advantage $MA = l_1/l_2 = F_2/F_1$: a long effort arm multiplies force but makes the effort point travel farther. In real levers part of the input work is lost to friction, so the efficiency $\eta = W_{out}/W_{in} = (F_2 d_2)/(F_1 d_1)$ is always below 1, and the ideal relation must be corrected to $F_1 l_1 \eta = F_2 l_2$.

A lever has an effort arm of 2.0 m and a load arm of 0.50 m. An effort force of 120 N is applied perpendicular to the long arm. Neglecting friction, find the load force that can be balanced and the mechanical advantage.

The same lever (effort arm 2.0 m, load arm 0.50 m) has an efficiency of 75%. What effort force is needed to raise a 480 N load? How much work does the effort do if the load rises by 0.10 m?

A worker must lift a 900 N crate with a lever whose bar is 1.50 m long. The effort must not exceed 200 N, and the efficiency of the lever is 75%. (a) Find the smallest effort arm for which this is possible. (b) Is the task feasible with this bar? (c) Compare the worker's work with the useful work if the crate rises by 0.05 m.

Run an investigation in a lever simulation (for example PhET Balancing Act or Torque) so that no risky loads are placed on real supports.

1. Place a 10 kg load 0.5 m to the left of the pivot. Choose a load on the right at 1.0 m that balances the beam, and record its mass.
2. Compute the left and right torques with $\tau = rF\sin\theta$ (weight $F = mg$, $\theta = 90^\circ$) and confirm they are equal.
3. Double the right-hand arm and repeat the balancing. Explain how the mechanical advantage $MA = l_1/l_2$ changed.
4. Set the efficiency to 70% (or switch on friction if available) and determine the force now needed to hold balance; compare the input and useful work.

Record your results in a table: arms, forces, torques, $MA$, $\eta$. Write a conclusion about the trade-off between force and displacement.

Show that the rotational effect of a force depends not only on the magnitude of the force but also on its lever arm, and that lever equilibrium means equal torques, not equal forces.

Diagram: a horizontal lever bar on a triangular fulcrum; to the left of the fulcrum an arrow for the load $F_2$ at distance $l_2$, to the right an arrow for the effort $F_1$ at distance $l_1$; the arms are marked with dashed lines and labelled $l_1$, $l_2$; curved arrows show the opposing torques $\tau_1 = F_1 l_1$ and $\tau_2 = F_2 l_2$. Below the diagram, a graph of the balancing force $F_1$ versus the effort arm $l_1$ at constant $F_2 l_2$: the hyperbola $F_1 = F_2 l_2 / l_1$, showing that a longer arm reduces the required force.

A worker pushes a wheelbarrow loaded with $50\ \text{kg}$ of sand. The load's centre of gravity is $0{,}40\ \text{m}$ from the wheel axle, and the hands grip the handles $1{,}20\ \text{m}$ from the same axle. The load's weight is $mg = 50 \times 9{,}8 \approx 490\ \text{N}$.

The lever equilibrium condition applies: the sum of torques about the axle is zero, $F_1 d_1 = F_2 d_2$. The torque of the load's weight, $490\ \text{N} \times 0{,}40\ \text{m} = 196\ \text{N·m}$, is balanced by the torque of the hands' force $F \times 1{,}20\ \text{m}$, so $F \approx 163\ \text{N}$ — a force advantage of exactly $1{,}20/0{,}40 = 3$.

Loosening a car wheel nut requires a torque of $90\ \text{N·m}$. The standard lug wrench has a lever arm of $0{,}30\ \text{m}$: producing that torque means pushing with a force $F = 90/0{,}30 = 300\ \text{N}$. Sliding a cheater pipe onto the wrench to extend the lever arm to $0{,}60\ \text{m}$ reduces the needed force to $150\ \text{N}$.

This is a direct use of the definition of torque, $\tau = F d$, where $d$ is the lever arm — the perpendicular distance from the axis of rotation to the line of action of the force. A force of $300\ \text{N}$ at $0{,}30\ \text{m}$ and a force of $150\ \text{N}$ at $0{,}60\ \text{m}$ give the same torque $90\ \text{N·m}$, because what matters is not the magnitude of the force alone but its product with the lever arm.

A block and tackle in a workshop lifts a $120\ \text{kg}$ engine through a height of $2{,}0\ \text{m}$. Its weight is $120 \times 9{,}8 = 1176\ \text{N}$. On the free end of the rope, a worker applies a force of only $300\ \text{N}$, but pulls $8{,}0\ \text{m}$ of rope through the system.

The pulley system gives a force advantage of about $1176/300 \approx 3{,}9$ times, but loses exactly the same factor in distance: $8{,}0/2{,}0 = 4$. The work input, $300\ \text{N} \times 8{,}0\ \text{m} = 2400\ \text{J}$, is almost equal to the useful work, $1176\ \text{N} \times 2{,}0\ \text{m} = 2352\ \text{J}$; the difference of $48\ \text{J}$ went into friction in the pulleys, so the efficiency is $\eta = 2352/2400 \approx 0{,}98$.