Question

A block of mass 10 kg is kept on a rought inclined plane as shown in the figure. A force of 3N is applied on the block. The coefficient of static fraction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g = 10 ms–2)



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Answer 1

Answer:

P = Pulling force

Angle of Inclination

Angle of friction

- wherein

For equilibrium

use

3 + mg sin

= P +

mg cos

P = 32 N

Explanation:

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Answer 2

Answer:

The answer to your question is 32 N.

Explanation:

Mass = 10 Kg

Force = 3 N

Static Friction = 0.6 N

g = 10 m/s²

So,

mg Sin45° = 100 / √2

Then we can say that,

μmg Cosθ = 0.6 × mg × 1 / √2

= 0.6 × 50√2

= 31.28 ≈ 32

So the force is 32 N.

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