A 2 kg block rests on a 34o incline. If the coefficient of static friction is 0.2, how much additional force, F, must be applied to keep the block from sliding down the incline?

Respuesta :

Answer:

Explanation:

Reaction force of inclined surface = mg cosθ

= Friction force acting in upward direction = μ x mg cosθ

If F be force required in upward direction to keep the block at rest on the plane

F +  μ x mg cosθ = mg sinθ

F = mg sinθ -  μ x mg cosθ

F = mg( sinθ - μ cosθ)

= 2 x 9.8 ( sin34 - 0.2 cos34 )

= 19.6 ( .559 - .1658)

= 7.7 N

This is the minimum force required .

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