A body of mass 2kg is kept on a rough horizontal surface. If μs=0.5 and an external force of F=5N is applied on the body, find the angle of friction. (Take g=10m/s2)
A
tan−1(12)
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B
tan−1(2)
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C
cot−1(2)
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D
cot−1(12)
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Solution
The correct option is Atan−1(12) The given situation can be shown as
The FBD of the block is
We have angle of friction as tanθ=fsN=μsmgmg=μs ⇒θ=tan−1(0.5)