The system shown is in limiting equilibrium. The coefficient of friction for all contanct surfaces is 1/4.

Options

A. `tan theta=(3)/(8)`
B. Tension in the string `=((100)/(3)g sin theta)N`
C. Net frictional force on `80 kg` block is `(80 g sin theta)N`
D. Force exerted by `20 kg` block on `80 g` block is `(20 g cos theta)`

Answer

A, B, C

Explanation

`20 g sin theta + f_(2) = T`
`20 g sin theta + mu(20 g cos theta) = T`
`80 g sin theta = mu(100 g cos theta) + mu (20 g cos theta)`
`tan theta = (3)/(8)`
`T = 20g sin theta + (1)/(4) xx 20 xx g xx (8)/(3) sin theta`
`= ((100)/(3) g sin theta) N`
Net friction on `80 kg =f_(1) + f_(2) = 80 g sin theta`
force on `80 kg` due to `20 kg` is
`sqrt((2 g cos theta)^(2)+(mu 20 g sin theta)^(2))`