# Ap Physics C 1989 M2?

## Jun 11, 2018

a) This is a basic application of newton's second law.

$2 M g - {T}_{v} = 2 M a$

$2 M g - 2 M a = {T}_{v}$

$2 M \left(g - a\right) = {T}_{v}$

b) First of all, we know that what will make the pulley have a non zero angular acceleration is the tension.

We know torque to be $I \alpha$ and $\alpha = \frac{a}{r}$, therefore, $\tau = I \left(\frac{a}{r}\right)$. Furthermore, torque is defined as $\text{lever arm" • "force}$, so we can say

${T}_{v} R - {T}_{h} R = I \left(\frac{a}{r}\right)$
${T}_{v} R - {T}_{h} R = 3 M {R}^{2} \left(\frac{a}{R}\right)$
${T}_{v} - {T}_{h} = 3 M a$
$2 g M - 2 M a - {T}_{h} = 3 M a$
$2 g M - 5 M a = {T}_{h}$
$2 g M - 5 M a = {T}_{h}$

c) Now for some linear dynamics! We know that the force of friction is given by ${F}_{f} = \mu {F}_{n} = \mu {m}_{\text{total}} g$

Our expression for net force will therefore be

${T}_{h} - {F}_{f} = 3 M a$
$2 M g - 5 M a - \mu {m}_{\text{total}} g = 3 M a$

Here we will be taking the mass of the top block for the normal force , because we are considering the friction between the two blocks.

$2 M g - 5 M a - \mu 4 M g = 3 M a$
$2 M g - 4 M g \mu = 8 M a$
$2 M g - 4 M g \mu = 8 M a$
$g - 2 g \mu = 4 a$

We know the value of $a$ so

$g - 2 g \mu = 4 a$
$g \left(1 - 2 \mu\right) = 8$
$1 - 2 \mu = \frac{8}{g}$
$\mu \approx 0.1$

d) The only force acting on the top box is friction, therefore

$4 M g \mu = {M}_{c} {a}_{c}$
$4 M g \mu = 4 M {a}_{c}$
$g \left(0.1\right) = {a}_{c}$
${a}_{c} \approx 1$ m/${s}^{2}$

This concludes this problem. Hopefully this helps!