Torque
Why where you push matters as much as how hard.
Identify the torques exerted on a rigid system. Describe the torques exerted on a rigid system.
Everything on this page serves that one sentence.
When you need to loosen a bolt that will not move, you reach for the longest wrench in the drawer. Nobody has to teach you that. What you may never have put into words is why it works, since your arm supplies the same force either way. Do you think the long wrench gives you more force? If not, what does it give you more of?
What it gives you is the quantity this topic is about. A force on its own does not tell you whether something will turn. Where that force is applied, and which way it points, matter just as much, and torque is the one number that holds all three together.
1. Where you push decides how much you turn it
When you push on a door at the handle, it swings easily. When you push with the same force right next to the hinges, almost nothing happens. The force has not changed and neither has the door, so the difference has to come from the distance between your hand and the hinge.
That distance is called the lever arm, and we measure it from the axis of rotation straight out to the line the force acts along. When you double the lever arm, you double the turning effect of the same force.
This is why the long wrench works. The bolt feels the same hand further out, and how far out the hand is turns out to be exactly what a bolt responds to.
A wrench on a bolt, seen from the side. Drag your hand along the handle to change how far out you push, and set the force with the slider. The lever arm is drawn as a labeled dimension line from the bolt to your hand, and the panel shows how hard you would have to push at the near end to produce the same torque.
Test thisSet the force to 40 N and put your hand at about \(r = 0.10\) m. Read the torque. Now drag your hand out to \(r = 0.30\) m without touching the force slider.
Report what happenedBy what factor did the torque change, and by what factor did the distance change? Explain in one sentence why those two factors have to agree.
Now without the simulationYou apply a perpendicular force of \(25\) N at a distance of \(0.40\) m from an axis. The torque you exert is
Topic questions
1 Two students push on the same door with equal forces, both perpendicular to the door. One pushes at the handle and one pushes halfway to the hinges. The one at the handle exerts
2 A mechanic cannot loosen a bolt with a short wrench. Which change increases the torque without a larger force from the mechanic?
2. Only the part of the force across the bar turns anything
When you pull on a door straight toward its hinges, the door does not swing at all, however hard you pull. The force is real and it is large and it does nothing, because none of it acts across the door.
Any force can be split into two parts. One part runs along the line from the axis to your hand, and the other runs perpendicular to that line. Only the perpendicular part turns anything, because the part along the line presses the system against its axle and the axle holds.
So the question is never only how hard you push. It is how hard you push across, which is the perpendicular part, and the simulation lets you watch the torque fall to zero while the force stays exactly where it was.
A bar hinged at one end. Drag the arrowhead to aim the force any way you like, and set its size with the slider. The force is split on screen into a part along the bar and a part across it, closed by a dashed rectangle, and the panel shows both parts next to the torque they produce.
Test thisSet the force to 40 N and aim it straight across the bar. Read the torque. Now turn the arrow until it points straight along the bar, toward the hinge.
Report what happenedWhat does the force read at the end, and what does the torque read? Explain what happened to the part of the force that was doing the turning.
Now without the simulationA force of \(30\) N acts at \(0.50\) m from an axis, at \(30^{\circ}\) to the line from the axis to the point where it acts. The torque is
Topic questions
3 A force is applied to a bar in a direction that points exactly along the bar, toward the axis. The torque it exerts about that axis is
4 Which arrangement produces the largest torque from a given force at a given distance from the axis?
3. The torque equation, and two ways to read it
When you put the lever arm and the perpendicular force together, you get the size of the torque.
\(\tau = rF_{\perp} = rF\sin\theta\)newton meters
Here \(\theta\) is the angle between the force and the line running from the axis out to the point where that force acts. When the two are perpendicular, \(\sin\theta\) is one and the torque is as large as that force at that distance can make it. When they lie along the same line, \(\sin\theta\) is zero and so is the torque.
There are two equally correct ways to read that product, and you should be comfortable with both of them. You can take the whole distance and the perpendicular part of the force, or you can take the whole force and the perpendicular distance from the axis to the line the force acts along. The two readings give the same number every time, and a problem will usually make one of them much easier than the other.
The unit is the newton meter. It is not a joule, even though a joule is also built from a newton and a meter, because a torque and an energy are different kinds of quantity that happen to share their units. Write N m for a torque and save the joule for energy.
Sliders for the distance, the force and the angle between them. The geometry is drawn on the left, with the perpendicular distance dropped from the axis onto the line of action, and on the right the torque is plotted against the angle from zero to one hundred eighty degrees, with your current angle marked on the curve.
Test thisSet \(r = 0.40\) m and \(F = 50\) N. Move the angle slider slowly from \(0^{\circ}\) up to \(180^{\circ}\) and watch the marked point travel along the curve.
Report what happenedAt which angle is the torque largest, and what is the torque at that angle? Explain what the curve is doing at zero degrees and at one hundred eighty degrees, and why the force is not zero at either of them.
Now without the simulationA force of \(20\) N acts \(0.60\) m from an axis, perpendicular to the line from the axis. If the angle is then changed to \(150^{\circ}\) with nothing else altered, the torque
Topic questions
5 The quantity \(r\sin\theta\) in the torque equation is best described as
6 A torque of 12 N m and an energy of 12 J have the same units written out. The reason they are still different quantities is that
4. Force diagrams keep the shape of the system
A free-body diagram was enough while the object was a point, because every force arrived at the same place. A rigid system needs more than that, since the same force at one end and at the other end do completely different things.
A force diagram keeps the shape of the system and draws every force at the point where it actually acts. Forces are labeled with their symbols, and the length of an arrow shows the relative size of a force rather than a value you are meant to measure off the page.
Once the diagram is drawn, each force gives one torque about whatever axis you have chosen, and you add those torques with their signs. Choosing a different axis changes the individual torques, which is allowed and often useful, because the axis you choose can make an inconvenient force give zero and disappear from the sum.
A beam with a pivot you can slide along it and three forces with their own sliders, one of them the weight of the beam acting at its center. Each force gets its own curved arrow showing which way that force alone would turn the beam, labeled with its signed torque, and the net cell turns green when the torques cancel.
Test thisSlide the pivot to the far left end and read the three torques. Now slide it to the far right end and read them again.
Report what happenedWhich torques changed sign as the pivot passed them, and which did not change sign at all? Explain in one sentence what decides the sign of a torque.
Now without the simulationA force diagram differs from a free-body diagram mainly because it
Topic questions
7 A student calculates the torques on a beam about an axis at its left end and gets a net torque that is not zero. A second student uses an axis at the right end and gets a different set of individual torques. The second student has
8 On a force diagram for a uniform beam, the weight of the beam is drawn
5. The sign of a torque says which way
A torque carries a sign for the same reason a velocity does. Counterclockwise is usually called positive, and whichever direction you choose, you maintain it for the whole problem.
When several torques act at once, the system responds to their sum. Two torques of equal size and opposite sign cancel, and the system does not turn, even though neither force has gone anywhere and both are still pressing on it.
Here is the point to be careful about. A system with no rotation is not a system with no forces. It is a system whose torques have added to zero, which is a much stronger and much more interesting statement, and Topic 5.5 is built on it.
Two forces on a disk. Drag each point around the rim and set each force with its own slider, with counterclockwise taken as positive. The panel shows each torque with its sign and their sum, and pressing Run turns the disk the way the net torque says it should.
Test thisSet the two forces so that the net torque reads zero, then press Run.
Report what happenedWhat does the disk do? Explain whether the two forces are still acting on it, and how you can tell from the figure.
Now without the simulationTwo torques act on a wheel: \(+8.0\) N m and \(-3.0\) N m. The net torque is
Topic questions
9 A rigid bar has two forces acting on it and does not rotate. It must be true that
10 Taking counterclockwise as positive, a force that would turn a wheel clockwise exerts a torque that is
Topic 5.4 asks the other half of the question. A torque is one side of it, and how hard the object is to turn is the other.