Question

Ap Physics C 1989 M2?

Answer 1

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

`2Mg - T_v = 2Ma`

`2Mg - 2Ma = T_v`

`2M(g - a) = 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 = a/r`, therefore, `tau = I(a/r)`. Furthermore, torque is defined as `"lever arm" • "force"`, so we can say

`T_vR - T_hR = I(a/r)`
`T_vR - T_hR = 3MR^2(a/R)`
`T_v - T_h = 3Ma`
`2gM - 2Ma - T_h = 3Ma`
`2gM - 5Ma = T_h`
`2gM-5Ma = 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_"total" g`

Our expression for net force will therefore be

`T_h - F_f = 3Ma`
`2Mg - 5Ma - mu m_"total" g = 3Ma`

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.

`2Mg -5Ma - mu 4Mg = 3Ma`
`2Mg - 4Mgmu = 8Ma`
`2Mg - 4Mgmu = 8Ma`
`g - 2gmu = 4a`

We know the value of `a` so

`g- 2gmu = 4a`
`g(1 - 2mu) = 8`
`1 - 2mu = 8/g`
`mu ~~ 0.1`

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

`4Mg mu = M_c a_c`
`4Mg mu = 4Ma_c`
`g(0.1) = a_c`
`a_c ~~ 1` m/`s^2`

This concludes this problem. Hopefully this helps!