First slide

Angle of repose

Question

A block of wood weights 10 N and is resting on an inclined plank. The coefficient of friction is 0.7. The frictional forces that acts on the block when the plank is 30° inclined with the horizontal is

Moderate

A

6.062N
B

5 N
C

9.8 N
D

70 N

Solution

See fig. of the question,
$From figure, R = 10 \cos 30^{\circ} = 5 \sqrt{3} N$
$and μR = 10 \sin 30^{\circ} = 5 N$
$(Note that μR = 0 \cdot 7 \times 5 \sqrt{3} = 6 \cdot 062 N$
This is maximum attainable frictional force. When the inclination of the plank is raised, then 10 sin $θ$ increases. The block will remain in rest till 10 sin $θ$ attains a value 6.062 N. After that the box will start sliding down the plank)

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