## Using force and torque to solve motion problems

When solving motion problems in physics, start by drawing a picture of the situation. Then use the picture to create an FBD. If there are no rotations, set the linear forces equal to $ma$ and solve. If there are rotations, set the linear forces equal to $ma$ and set all the torques equal to $I \alpha$. Then solve the equations.

Example 1: A block of mass $0.50 \, kg$ is attached to the end of a string of negligible mass wrapped around a wheel of radius $0.30 \, m$ as shown. If the wheel has a moment of inertia of $0.18 \, kg \cdot m^2$, find the acceleration of the mass as it falls.

We can find all the information we need by writing the linear and rotational equations of force. The block is experiencing only linear motion, so there is one equation for it. The two forces acting on it are gravity and the tension in the string:

$mg-T=ma$

The wheel is experiencing only rotational motion, so there is only one equation for it too. The only torque acting on the wheel is due to the tension in the string:

$\tau=F \cdot r=Tr=I \alpha$

It appears at first as though there are too many unknowns: $T, a, \text{ and } \alpha.$ But remember that $a=r \alpha.$ So we can write the second equation as:

$Tr=I \alpha =I \dfrac{a}{r}$

Solve both equations for $T$ and substitute:

$mg-ma=I \dfrac{a}{r^2} \rightarrow mg=I \dfrac{a}{r^2}+ma$

$\therefore a= \dfrac{mg}{\dfrac{I}{r^2}+m}= \dfrac{0.50 \cdot 9.8}{\dfrac{0.18}{0.30^2}+0.50}=2.0 \, m/s^2$

## When is torque useful in an equilibrium problem?

When you are trying to find a particular force for a system in equilibrium, you draw an FBD and set the sum of all the forces equal to zero. But sometimes, this doesn’t provide enough information for you to calculate the missing value. For example, check this problem out:

Example: A 6 m bench is supported by two posts at its two ends. The bench is a uniform piece of wood with a weight of 200 N. A dog with a weight of 100 N sits 2 m from one end of the bench. Find the magnitudes of the forces exerted on the bench by the two supports.

Solution: If we start by drawing a schematic of the bench, we can attach all the forces:

Obviously, the four forces sum to zero—the two support forces pushing up balance the weight of the bench and the dog. But that only gives us one equation and we have two unknowns—the forces upward by each leg of the bench.

How do we solve the problem? We recognize that the sum of all the torques also equals zero. But if the bench is not rotating, what’s the pivot point? We can use any point we want! Let’s pick the left end of the bench, because then the force from the left support does not exert a torque. There are then two clockwise torques (the center of mass of the bench and the dog) and one counterclockwise torque (the right leg of the bench). Now we have two equations:

$F_1+F_2-F_{bench}-F_{dog}=F_1+F_2-200-100=0$

$\tau_{bench}+\tau_{dog}-\tau_2=(200)(3)+(100)(4)-(F_2)(6)=0$

$\therefore F_2=167 \, N \text{ and } F_1=133 \, N$

Note that we could have chosen any other point along the bench as our pivot point, and we would have obtained the same result.

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