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The situation here is similar to that depicted above, though the distances of the rubber bands from the center are different.  The die on which the strap rotates is not visible, but is attached to the tabletop; the blue push pin constrains the system to rotate about a vertical axis through the center of the die.

The problem:

The metal strap used in the Angular Velocity of a Strap experiment is constrained by a vertical push pin to rotate about a hole in a die.  The die is glued in place to a massive tabletop.  A rubber band is attached to a point 15 cm from the axis and stretched so that it exerts a force of 3 Newtons, directed perpendicular to the rod.  If this force is unopposed it will accelerate the system rapidly. You want to attach a second rubber band 5 cm from the axis to prevent the system from rotating. 

• How much force will that rubber band have to exert? 

The most intuitive idea is that the rubber band at 15 cm has three times the leverage of the rubber band at 5 cm, so to have the same effect the second must exert three times the force.

The first force is said to apply a torque (a torque is sort of a twisting force, actually determine by the force and its moment-arm or leverage) equal to .15 m * 3 N = .45 m N.

To balance the torque or twisting effect first, the second force must exert an equal and opposite torque. If F is the second force, then its torque is .05 m * F.

To say that the torques are equal and opposite is to say that .45 m N = - .05 m * F, so that F = -.45 m N / (.05 m) = -9 N.

• The negative sign indicates that the second force is directed so that it rotates the system in the direction opposite the first.

• m N not Joules; no work here; torque interpretation

Note that work is equal to force * displacement along the line of the force. In this case nothing is moving so there is no displacement, so no work is involved here.

However the units are identical to those of work, which is a frequent source of confusion.

In fact what this product gives you it torque:

• 3 N is the force, .15 m is called the moment-arm of the force, and the torque is the product of moment-arm and force: .15 m * 3 N = .45 m * N.

• Note that I've expressed this as m * N to indicated that this is a torque and not work; in fact m * N and N * m are mathematically identical units with two different physical interpretations, the choice of m * N for torque or N * m for work is more or less a grammatical convention.  However we do need to be careful to call this quantity 'torque', and not confuse it with work.  The 'grammatical convention' is useful to help remind us whether the quantity represents work or force.

• When used to refer to torques, m * N (or N * m) does not equal Joules. Joules measure work, not torque.

Here we say that since the system is in equilibrium (not accelerating) the torques must be equal and opposite, so that | .15 m * 3 N | = | .05 m * F |, where F stands for the unknown force.

• The solution is | F | = .45 m N / (.05 m) = 9 N.

There are also conventions for signs of forces and torques, which you might already have encountered.  These conventions which will be developed within the next few assignments.

• Once both of these forces are in place and the system is stationary, what (if anything) will happen if the glue holding the die to the tabletop comes loose?

The net force on the system is 12 N, the sum of the forces of the two rubber bands.  If the glue comes loose, it isn't unreasonable to expect that the system will accelerated in the direction of the net force.  This isn't actually the case, however. 

• The inequality of the forces will in fact cause some of the energy of the rubber bands to go into the rotation of the system.

• Note also that the forces exerted by the rubber bands begin to change as they contract.

However as long as the glue holds, the pin through the axis of rotation would exert a force of 12 N.