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 kg block is `(20 g cos theta)`

Answer

`A, B, C`

Explanation

`20gsintheta+f_(2)=T`
`20g sintheta+mu(20g costheta)=T`
`80gsin theta=mu(100gcostheta)+mu(20gcostheta)`
`tantheta=(3)/(8)`
`T=20g sintheta+mu20gcostheta`
`=20g sintheta+(1)/(4)xx20xxgxx(8)/(3) sintheta`
`((100)/(3)gsintheta)N`
Net friction on `80kg =f_(1)+f_(2)=80 gsintheta`
force on`80kg` due to `20kg` is `sqrt((20g costheta)^(2)(mu20gsin theta)^(2))`