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Question

Two blocks of masses m1 and m2 are connected through a massless inextensible string. Block of mass m1 is placed at the fixed rigid inclined surface while the block of mass m2 hanging at the other end of the string, which is passing through a fixed massless frictionless pulley shown in figure. The coefficient of static friction between the block and the inclined plane is 0.8. The system of masses m1 and m2 is released from rest. (Take g=10 m/s2)

A
The tension in the string is 20 N after releasing the system
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B
The contact force by the inclined surface on the block is along normal to the inclined surface
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C
The magnitude of contact force by the inclined surface on the block m1 is 203N
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D
Frictional force between 4 kg and inclined plane is zero
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Solution

The correct option is D Frictional force between 4 kg and inclined plane is zero
Net force on the system of masses is m1gsin30m2g=4gsin302g=0

Since net pulling force is zero on the system, static friction adjusts its value to zero
i.e f=0
and net contact force by the incline is perpendicular to the incline.
N=m1gcos30
For m2
Tm2g=0
Putting a=0 and f=0
T=20 N
Why this question?

Tip: Since static friction is self adjusting, we should first find the resultant of all the forces parallel to the interface of the bodies, and then one will get an idea as to how much friction is required, or it is even required or not!

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