An auto-rickshaw has a flat tyre on a busy road in Jaipur. The driver tries to loosen a wheel nut with a short spanner and cannot move it. A mechanic arrives, slips a steel pipe over the handle and pushes down on the far end. The nut turns at once. The mechanic did not push harder than the driver. He pushed farther from the nut, and straight across the handle.
Where and how a force acts
Push a door at its handle and it swings open easily. Push with the same force close to the hinge, and the door barely moves. Push straight along the door, towards the hinge, and it does not turn at all. The size of the force was the same each time. What changed was where the force acted and in which direction.
A door is a rigid body that turns about a fixed line, its axis of rotation. The quantity that measures how strongly a force turns a body about an axis is the torque. It is written \(\tau\), the Greek letter tau. Torque is to rotation what force is to translation: a net torque changes the angular velocity of a body.
Look at Fig 1.1. At the handle, at right angles to the door, the push gives \(16\,\text{N m}\). At the middle of the door the same push gives only \(8\,\text{N m}\), half as much. At the handle but at \(30°\) to the door it also gives \(8\,\text{N m}\). Pulled straight along the door, it gives no torque at all.
Torque about an axis
Let a force \(\vec F\) act at a point P of a body that turns about a fixed axis through O. Let the force lie in the plane that is perpendicular to the axis. Let \(r\) be the distance of P from the axis, and let \(\theta\) be the angle between the line OP and the force. Then the torque of the force about the axis is \[ \tau = rF\sin\theta \]
The torque of a force about an axis is \(\tau = rF\sin\theta\). Here \(F\) is the force in the plane perpendicular to the axis, and \(r\) is the distance of its point of application from the axis. The angle \(\theta\) is measured between the force and the line drawn from the axis to that point.
Torque carries a sign. Seen from one side of the axis, a force may turn the body anticlockwise or clockwise. This book takes anticlockwise torques as positive and clockwise torques as negative, the same convention as for angles. Two torques in opposite senses can then be added with their signs.
The unit of torque is the newton metre, N m. Work is also force times distance, and the joule is also a newton metre. Yet torque is never quoted in joules. Work uses the distance moved along the force. Torque uses a distance measured across the force, from the axis. They are different quantities that happen to share a unit.
A rope tied to the edge of a door is pulled so that its line passes exactly through the hinge line. The door is \(0.8\,\text{m}\) wide and the pull is \(30\,\text{N}\). What is the torque about the hinge?
Only the part across the line turns
Why does \(\sin\theta\) appear in the torque? Resolve the force at P into two parts. One part, \(F\cos\theta\), points along the line OP. It is the radial component. It pulls P towards the axis or pushes it away, and a rigid body cannot let P move that way. The other part, \(F\sin\theta\), points across the line OP. It is the transverse component, and it moves P round the axis.
So only the transverse component turns the body. Its effect also grows with the distance \(r\) at which it acts. Double \(r\) and the same transverse push turns the door twice as strongly. The torque is therefore \(r\) times \(F\sin\theta\), the transverse component acting at distance \(r\).
A force can also have a part parallel to the axis itself, such as a push along a door's hinge line. That part slides the body along the axis, or strains the hinges, but it cannot turn the body about the axis. So only the components in the plane perpendicular to the axis enter the torque.
Two forces act at right angles to a door. One is \(24\,\text{N}\) at \(0.1\,\text{m}\) from the hinge. The other is \(6\,\text{N}\) at \(0.5\,\text{m}\) from the hinge. Which has the larger torque about the hinge?
The second way: the moment arm
The product \(rF\sin\theta\) can be grouped in a second way, as \(F\) times \((r\sin\theta)\). This grouping has its own picture. Extend the force in both directions to get its line of action. From the axis, drop a perpendicular onto that line. Its length is \(d = r\sin\theta\), called the moment arm or lever arm. The torque is the whole force times its moment arm. \[ \tau = r\,(F\sin\theta) = F\,(r\sin\theta) = F\,d \]
The second way has one useful result. Slide a force along its own line of action, and its moment arm does not change. So its torque does not change either. It does not matter where on the line the force is applied, only where the line lies. A rope pulling a gate turns it equally, whether it is tied near the gate or far out along the same line.
A spanner is gripped \(0.3\,\text{m}\) from the nut. A force of \(50\,\text{N}\) acts there, at \(30°\) to the handle. Find the torque about the nut by the transverse component and by the moment arm.
\(r = 0.3\,\text{m}\), \(F = 50\,\text{N}\) and \(\theta = 30°\), so \(\sin\theta = 0.5\).
First way: \(\tau = r\,(F\sin\theta)\). Second way: \(\tau = F\,(r\sin\theta)\).
Transverse component \(F\sin\theta = 50 \times 0.5 = 25\,\text{N}\), so \(\tau = 0.3 \times 25 = 7.5\,\text{N m}\). Moment arm \(d = 0.3 \times 0.5 = 0.15\,\text{m}\), so \(\tau = 50 \times 0.15 = 7.5\,\text{N m}\).
The torque is \(7.5\,\text{N m}\) either way. With components given, the first way is quicker. With the line of action drawn on a diagram, the second way is quicker.
A force of \(12\,\text{N}\) acts at one end of a rod \(0.5\,\text{m}\) long, at \(60°\) to the rod. What is its torque about an axis through the other end, perpendicular to the rod and the force, in N m?
Getting the most torque
Move the slider in the figure above from \(0°\) to \(180°\). At \(0°\) the force points along the handle, its line passes through O, and the torque is zero. The torque then rises, reaches \(15\,\text{N m}\) at \(90°\), and falls to zero again at \(180°\). For a fixed \(r\) and \(F\), the torque is largest when \(\sin\theta = 1\). So push at right angles to the handle.
The other factor is the distance. At \(90°\), \(\tau = rF\), so the force needed for a given torque is \(F = \tau/r\). Suppose a wheel nut needs \(90\,\text{N m}\) to loosen. With a spanner \(0.3\,\text{m}\) long, the push must be \(90/0.3 = 300\,\text{N}\). With a pipe that makes the handle \(0.6\,\text{m}\) long, it is only \(150\,\text{N}\).
The angle \(\theta\) in \(rF\sin\theta\) is measured between the force and the line from the axis to the point of application. It is not the angle with the floor or with the wall. When a question gives another angle, draw the line from the axis first and measure from it.
Four forces act on a door at different points, all about the same hinge. Rank their torques from the largest to the smallest.
A cyclist in Pune stands on one pedal of a bicycle going uphill, with the crank level. Estimate the torque about the crank axle, in N m.
Explain why the mechanic's pipe loosens a wheel nut that a short spanner could not, although the mechanic pushes no harder.
The nut turns when the torque about it is large enough. At right angles the torque is the force times the distance from the nut, \(rF\). The pipe doubles the distance, so the same push gives twice the torque. Pushing at right angles makes \(\sin\theta = 1\), so none of the push is wasted along the handle.
The Greek mathematician Archimedes studied the lever, a rigid bar turning on a fixed support. He showed that two weights balance when each weight times its distance from the support is the same on both sides. That product is the torque of each weight. He is said to have boasted that, given a place to stand, he could move the Earth. The boast is a claim about moment arms: with a long enough lever, a small force can give any torque.
Now return to the auto-rickshaw. The nut needed some torque, perhaps \(90\,\text{N m}\), before it would turn. The driver's short spanner needed a push of \(300\,\text{N}\) at right angles, more than he could give. The mechanic's pipe doubled the moment arm, so \(150\,\text{N}\) was enough. By pushing straight across the pipe, he made sure that all of the push turned the nut.
A car's wheel nuts must be tight, but not so tight that the bolts stretch. So a service centre uses a torque wrench. It has a spring inside, and it clicks or shows a reading when the torque reaches the value set on it. The mechanic pushes at right angles on the marked grip, where the tool is calibrated. The number printed in the service manual is a torque in N m, not a force.
- Using the angle between the force and the floor or the wall in \(rF\sin\theta\).Draw the line from the axis to the point of application, and measure \(\theta\) between it and the force.
- Using \(\cos\theta\) because the angle is given with the rod.The turning part is across the rod, \(F\sin\theta\). The part along the rod, \(F\cos\theta\), turns nothing.
- Writing torque in joules.Torque is in N m. Its distance is measured across the force; work's distance is measured along it.
- Taking the distance to the point of application as the moment arm.The moment arm is the perpendicular distance from the axis to the line of action, \(d = r\sin\theta\).