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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))`

A

`150 N`

B

`75N`

C

`200 N`

D

`250 N`

Text Solution

Verified by Experts

The correct Answer is:
a

If the plane as an angle `theta` with horizontal then can `theta = 8//15` If R is the normal.
`R= 170 kg cos theta = 170 xx 10 xx((15)/(17)) = 1500 N`
Force of friction on `A = 1500 xx 0.2 = 300N`
Force of friction on `B = 1500 xx 0.4 = 600N`
Considering the two blocks as a system the netforce paraller to the plane is
`= 2 xx 170g sin theta - 300 - 600 = 1600 - 900 = 700 N`
Acceleration `= (700)/(340)= (35)/(17)ms^(-2)`
Consider the motion of A along
`170g sin theta - 300 - P= P170 xx (35)/(17)`(where P is pull on the bar)
`P = 500 - 350 = 150 N`
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