Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.
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Answer:
Explanation:
g=10ms2
m=2kg
θ=300,m,u=0.2
From R−mgcosθ−Fsinθ=0
→R=mgcosθ+Fsinθ……..i
and mgsinθ+μR−Fcosθ=0
→mgsinθ+μ(mgcosθ+Fsinθ)−Fcos
θ=0
→mgsinθ+μmgcosθ+μFsinθ−Fcosθ=0
→F=(mgsinθ+μcosθ)μsinθ−cosθ
=(2×10×(12)+0.2×2×10×3√2)0.2×(12)−(3√2)
=13.4640.76=17.7N=17.5N
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