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Two blocks of masses m(1) and m(2) are c...

Two blocks of masses `m_(1)` and `m_(2)` are connected by a string of negligible mass which pass over a frictionless pulley fixed on the top of an inclined plane as shown in figure. The coefficient of friction between `m_(1)` and plane is `mu`.

A

If `m_1=m_2` the mass `m_1` first begin to move up inclined plane when the agle of inclination is `theta`,then `mu=tantheta`.

B

If `m_1=m_2` the mass `m_1` first begin to move up the inclined plane when the angle of inclination is `theta`, then `mu=sectheta-tantheta.

C

If `m_1=2m_2` the mass `m_1` first begin to slide down the plane if `mu=2tantheta`.

D

If `m_1=2m_2` the mass `m_1` first begin to slide down the plane if `mu=tantheta-(1)/(2)sectheta`.

Text Solution

Verified by Experts

The correct Answer is:
B, D
(b) The blocks `m_1` will just begin to move up the plane if the downward force `m_2g` due to mass `m_2` trying to pull the mass `m_1` up the plane just equals the force
`m_2g=(m_1gsintheta+mum_1gcostheta)`
`mg=mgsintheta+mucostheta`
`mg(1-sintheta)=mumgcostheta`
`mu=sectheta-tantheta`
So, choice (b) is correct
(d) The block `m_1` will just begin to move down the plane, if the downward force `(m_1gsintheta-mum_1gcostheta)` on `m_1` just equals the upward force `m_2` of acting on `m_1` due to `m_2` if
`m_2g=m_1g(sintheta-mucostheta)`
`m_2g=2m_2g(sintheta-mucostheta)`
`(1)/(2)=sintheta-mucostheta`
`(1)/(2costheta)=tantheta-mu`
`mu=tantheta-(1)/(2)sectheta`
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