First slide

Friction on inclined plane of angle more than angle of repose

Question

A block of mass 2 kg is lying on an inclined plane, inclined to the horizontal at 30°. If the coefficient of friction between the block and the plane is 0.7 then magnitude of frictional force acting on the block will be

Moderate

A

11 .9 newton
B

1.19 newton
C

0.19 newton
D

1109 newton

Solution

See fig.
$\begin{aligned} Here, F & = μR = μ \cdot mgcos θ = μ \cdot mgcos 30^{\circ} \\ = 0 \cdot 7 \times 2 \times 9 \cdot 8 \times (\sqrt{3} / 2) = 11 \cdot 9 N \end{aligned}$

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