First slide

Friction

Question

A block of mass m = 2 kg is resting on a rough inclined plane of inclination $37^{0}$ . The coefficient of friction between the block and the plane is $μ$ = 0.5. What minimum force F (in newton) should be applied perpendicular to the plane on the block, so that the block does not slip on the plane?

Moderate

Solution

Since mg sin 37°> $μ$ mg cos 37°, the block has a tendency to slip downwards.
Let F be the minimum force applied on it, so that it does not slip. Then,
$\begin{array}{l} N = F + mgcos 37^{\circ} \\ ∴ mgsin 37^{\circ} = μN = μ (F + mgcos 37^{\circ}) \end{array}$
$\begin{array}{l} or F = \frac{mgsin 37^{\circ}}{μ} - mgcos 37^{\circ} \\ = \frac{(2) (10) (3 / 5)}{0 .5} - (2) (10) (\frac{4}{5}) = 8 N \end{array}$

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