A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. Calculate the coefficient of friction between the block and inclined plane. Consider g=10ms−2
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Solution
Since the string is under tension, there is limiting friction acting between the block and the plane.
∑Fx=0⇒50+μN cos 45 = N cos 45 (1−μ)N√2=50....(i) ∑Fy=0⇒μNcos45+Ncos45=150 (1+μ)N√2=150......(ii)
Equation (ii)/(i)⇒1+μ1−μ=15050 1+μ=3−3μ⇒4μ=2μ=12