A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is [Take g=10 m…

Question

A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is [Take g=10 m/s²]

Accepted Answer

(2) (√3)/2. Idea: two limiting cases, two equations, two unknowns — the weight W and the friction μ W cos θ. Adding and subtracting them separates the two neatly. Pushed down with 2 N, on the verge of sliding down, so friction acts up at its maximum: W sin θ+2=μ W cos θ ...(i) Pushed up with 10 N, on the verge of sliding up, so friction acts down at its maximum: 10=W sin θ+μ W cos θ ...(ii) Adding (i) and (ii), the W sin θ terms cancel one way: 12=2μ W cos θ, so μ W cos θ=6 N. Subtracting, 8=2W sin θ, so W sin θ=4 N. With θ=30°: W=4/(sin 30°)=8 N, and W cos θ=8×(√3)/2=4√3. μ=6/(4√3)=3/(2√3)=(√3)/2. Note μ=0.87>tan 30°=0.58, which is consistent: the block sits happily on the slope and has to be pushed to move either way.

Step-by-step solution

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