torque lever arm and force composition

A rigid body cannot be reduced to a single point: a force applied to it not only accelerates the centre of mass but also tends to rotate the body about an axis. The measure of this rotational effect is the torque about a chosen point $O$: $M_O = F\,d,$ where $d$ is the lever arm, i.e. the length of the perpendicular dropped from $O$ to the line of action of the force. If the angle $\alpha$ between the force vector $\vec F$ and the position vector $\vec r$ from $O$ to the point of application is known, then $d = r\sin\alpha$, and in vector form $\vec M = \vec r \times \vec F$. Torque is taken as positive for counterclockwise rotation and negative for clockwise; the only requirement is to follow one stated convention consistently. Equilibrium of a rigid body requires not only $\sum \vec F = 0$ but also $\sum M_O = 0$ about any point. Forces applied to a body can be added vectorially: the resultant $\vec R = \sum \vec F_i$ replaces them in the equation of motion of the centre of mass, but it does not necessarily replace them with respect to rotation.

A force $F = 50\ \text{N}$ is applied to a wrench at a point located at a distance $r = 0{,}40\ \text{m}$ from the axis of the nut. The angle between the force vector and the segment from the axis to the point of application is $\alpha = 30^\circ$. Find the torque about the axis of the nut.

A weightless rod is fixed on an axis $O$. On the left, at a distance $0{,}20\ \text{m}$ from the axis, a force $F_1 = 40\ \text{N}$ is applied perpendicularly downward. On the right, at a distance $0{,}50\ \text{m}$, a force $F_2$ is applied at an angle $30^\circ$ to the rod. 1) Find the $F_2$ at which the rod is in equilibrium. 2) Find the magnitude of the force exerted on the rod by the axis.

Two forces of equal magnitude $F = 15\ \text{N}$ are applied to a weightless rod of length $L = 0{,}60\ \text{m}$: downward at the left end and upward at the right end; both are perpendicular to the rod. 1) Find the resultant of all forces. 2) Compute the total torque about the midpoint of the rod and about the left end. 3) Explain why knowing only the resultant is not enough to describe the behaviour of the rod.

Experiment with a ruler and coins. Place a ruler on a round pencil so that it balances: the pivot point is the axis of rotation. Put identical coins on the left and on the right at different distances from the pivot and find a position where the ruler stays horizontal; measure the lever arms and verify that $\sum M = 0$ (the torque of one coin is proportional to its distance from the pivot). Then tie a thread to the ruler and pull at $90^\circ$, and then at $30^\circ$ with the same force — the torque decreases, because the lever arm is $r\sin\alpha$. If you have no ruler or coins, open the PhET "Balancing Act" simulation and repeat the same experiments with weights on a beam, including the case of two equal opposite forces (a couple). Work on a table, do not suspend loads over the table edge, and do not use heavy or sharp objects.

Show that torque depends not on the force or the distance separately but on the lever arm — the perpendicular to the line of action of the force — and that torques add as signed quantities while forces add as vectors.

Diagram: a horizontal rod with axis $O$; at point $A$ a force $\vec F$ is applied at an angle $\alpha$ to the rod; a dashed perpendicular of length $d = r\sin\alpha$ is dropped from $O$ to the line of action of the force. Beside it, a vector triangle of force addition $\vec F_1 + \vec F_2 = \vec R$ and a small table of torque signs (counterclockwise — "$+$", clockwise — "$-$").

On a school seesaw, a child of mass $30$ kg sits $1.5$ m left of the pivot, and a child of mass $45$ kg sits $1.0$ m right of it. Their weights are $294$ N and $441$ N. The left torque is $294\,\mathrm{N}\cdot1.5\,\mathrm{m}=441\,\mathrm{N}\cdot\mathrm{m}$; the right torque is $441\,\mathrm{N}\cdot1.0\,\mathrm{m}=441\,\mathrm{N}\cdot\mathrm{m}$ in the opposite direction. The board stays horizontal.

This uses torque about an axis: the turning effect of weight depends on the distance to the pivot. Opposite signs of the torques show how rotational effects combine; the forces are vertical, and their sum is balanced by the pivot reaction force.

A driver loosens a wheel nut with a wrench whose handle is $0.30$ m long. If the driver pushes perpendicular to the handle with a force of $150$ N, the torque is $150\,\mathrm{N}\cdot0.30\,\mathrm{m}=45\,\mathrm{N}\cdot\mathrm{m}$. If the force is directed at $45^\circ$ to the handle, the perpendicular component is $150\sin45^\circ\approx106\,\mathrm{N}$, and the torque is $106\,\mathrm{N}\cdot0.30\,\mathrm{m}\approx31.8\,\mathrm{N}\cdot\mathrm{m}$.

This scenario trains resolution of a force into components and the lever-arm idea: only the component perpendicular to the line from the axis to the point of application creates torque. The lever arm can be taken as $d=r\sin\theta$, where $r=0.30$ m, $\theta=45^\circ$.

The width of a refrigerator door from the hinges to the handle is $0.60$ m. If you pull the handle with a force of $15$ N perpendicular to the door plane, the torque about the hinge axis is $15\,\mathrm{N}\cdot0.60\,\mathrm{m}=9\,\mathrm{N}\cdot\mathrm{m}$. To obtain the same torque by pushing the door $0.10$ m from the hinges, a force of $9/0.10=90$ N is needed. If you push along the door plane toward the hinges, the torque is zero.

This shows that torque is determined by the lever arm — the shortest distance from the axis to the line of action of the force. A force whose line of action passes through the axis has zero lever arm; a force perpendicular to the door uses the full lever arm $0.60$ m.