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Question

A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block as shown in the figure. The block remains stationary if (take g=10 m/s2)

A
θ=45
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B
θ>45, and a frictional force acts on the block towards P
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C
θ>45, and a frictional force acts on the block towards Q
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D
θ<45, and a frictional force acts on the block towards Q
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Solution

The correct option is C θ>45, and a frictional force acts on the block towards Q
From the FBD of small block, while at rest
Along the plane,
sin(90θ)=mgsinθfs
1cosθ=mgsinθμR ... (1)
Perpendicular to the plane,
R=mgcosθ+cos(90θ)
R=mgcosθ+1sinθ ... (2)
From (1) & (2) we get
cosθ=mgsinθμ×mgcosθμsinθ
cosθ+μsinθ=mg(sinθμcosθ)
cosθ+μsinθsinθμcosθ=mg1=1
For a frictionless surface μ=0
θ=45

Since for θ=45 no friction is required to keep the block stationary, therefore as the value of θ becomes more than 45 , the component of gravitational force (mgsinθ) increases and friction acts towards Q to keep the block stationary.


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