A small block of mass of 0.1kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1N acts on the block as shown in the figure. The block remains stationary if (take g=10m/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 Cθ>45∘, and a frictional force acts on the block towards Q From the FBD of small block, while at rest Along the plane, sin(90−θ)=mgsinθ−fs ⇒1cosθ=mgsinθ−μR ... (1) Perpendicular to the plane, R=mgcosθ+cos(90−θ) R=mgcosθ+1sinθ ... (2) From (1) & (2) we get cosθ=mgsinθ−μ×mgcosθ−μsinθ cosθ+μsinθ=mg(sinθ−μcosθ) cosθ+μsinθsinθ−μcosθ=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.