## 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:

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

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

## Pulley problems

In a pulley problem, if you have a chain of boxes all connected by ropes, first find the tension in the rope that goes over the pulley (treat all of the boxes on the table as a single box and all of the boxes hanging down as a single box). Then work your way to each end of the chain one box at a time to find the tension in each rope along the chain.

Example: Find the tensions in each cable and the acceleration of the blocks in the diagram below. Assume that the pulley is massless and the table and pulley are frictionless.

Solution: First, treat the two blocks on the table as a single unit of 5 kg, and treat the two hanging blocks as a single unit of 5 kg:

The net force equation for the “block” on the table is:

T2 = ma = 5a                                (1)

The net force equation for the hanging “block” is:

mg – T2 = ma → 5g – T2 = 5a             (2)

Plugging equation (1) into equation (2) gives

a = g/2 = 4.9 m/s2 and T2 = 5g/2 = 24.5 N.

Now you can find the other tensions. Find T3 by doing an FBD on the 3 kg block:

3g – T3 = 3a = 3g/2

Solve to get T3 = 3g/2 = 14.7 N. Find T1 by doing an FBD on the 4 kg block:

T1 = 4a = 2g, so T1 = 19.6 N.

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