Question

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The magnitude of the force is different in the two cases, while the directional angle θ is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is 0.310. The weight of the block is 55.0 N, and the directional angle for the force Upper F Overscript right-arrow EndScripts is θ = 31.0°. Determine the magnitude of Upper F Overscript right-arrow EndScripts when the block slides (a) up the wall and (b) down the wall.

Show all steps please!

Answer 1

For block sliding up the wall, both weight and force of kinetic friction must act in the same direction

as applied force is at an angle, we need to break it into two components.

The normal component will be Fsin

The vertical component will be Fcos

so,

using Newton's second law (remember, the question mentions constant velocity means a = 0 )

F_net = ma

F_net = 0

Fcos = W + F_kinetic

Fcos = W + uFsin

F ( cos - u sin) = W

F = W / cos - u sin

just put the values

F = 55 / (cos 31 - 0.310 * sin 31 )

F = 78.85 N

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for block moving down the wall, friction will act upwards,

so,

Fcos + F_kinetic = W

F = W / cos + u sin

F = 55 / (cos 31 + 0.310 * sin 31 )

F = 54.08 N