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Question

A body of mass 2 kg requires a force 3F to just move it up an inclined plane of angle 37 and a force F to prevent it from sliding down the same plane. Find the coefficient of friction. Take g=10 m/s2.

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Solution

The FBD of the block in both situations is shown below.


From the FBDs, we have
N=mgcos37 and
f=μN=μmgcos37 (limiting friction in both cases)
For case (1):
3F=mgsin37+μmgcos37(1)
For case (2):
mgsin37=F+μmgcos37(2)
Substituting F from eq. (2) in (1),
3(3mg5μmg×45)=35mg+45μmg
912μ=3+4μ 16μ=6
μ=616=38=0.375

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