Blocks A and B in the Fig are connected by a bar of negligible weight .Mass of each block is `170 kg` and `mu_(A) = 0.2` and `mu_(B) = 0.4` where `mu_(A)` and `mu_(B)`are the coefficient of limiting friction between bloock and plane calculate the force developed in the bar`(g = 10 ms^(-2))`

Options

A. `150N`
B. `75N`
C. `200N`
D. `250N`

Answer

A

Explanation

If the plane makes an angle theta with horizontal
`tantheta=8//15` If `R` is the normal reaction
`R=170gcostheta=170xx10xx((15)/(17))=1500N`
Force of friction on `A=1500xx0.2=300N`
Force of friction on `B=1500xx0.4=600N`
Considering the two block as a system, the net force parallel to the plane
`=2xx170gsintheta-300-600=1600-900=700N`
:. Acceleration `=(700)/(340)=(35)/(17)m//s^(2)`
Consider the motion of `A` alone.
`170gsintheta-300-P=170xx(35)/(17)`
(where `P` is pull on the bar).
`P=500-350=150N` .