A body of mass $10 k g$ lies on a rough inclined plane of inclination $θ = 37^{\circ}$ with the horizontal. When the force of $30 N$ is applied on the block parallel to and upward the plane, the contact force by the plane on the block is nearby along

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Solution

The correct option is A OA
The given arrangement can be reframed as shown

So, the force down the incline is
$m g sin θ = 10 g sin 37^{\circ} = 100 \times (\frac{3}{5}) = 60 N$
This $60 N > 30 N$ which is applied on the block externally.So, the block will slide down the plane.Hence, friction force $f$ will act up the incline as shown in the below figure.

Thus, contact force by the plane on the block is given by
$R = \sqrt{N^{2} + f^{2}}$
And, the direction will be near by along $O A$ as per given in the question.

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Q. A body of mass

10 k g

lies on a rough inclined plane of inclination

θ = 37^{\circ}

with the horizontal. When the force of

30 N

is applied on the block parallel to and upward the plane, the contact force by the plane on the block is nearby along

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10 k g

lies on a rough inclined plane of inclination

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with the horizontal. When the force of

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is applied on the block parallel to an upward the plane, the total force by the plane on the block is nearly along:

A body of mass $10 kg$ lies on a rough inclined plane of inclination $θ = {sin}^{- 1} \frac{3}{5}$ with the horizontal. When a force of $30 N$ is applied on the block parallel to and upward the plane, the total reaction by the plane on the block is nearly along (take $g = 10 m s^{- 2}$ )