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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 is θ = 50.0°. Determine the magnitude of when the block slides (a) up the wall and (b) down the wall.

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Answer:

Part a)

[tex]F = 135.7 N[/tex]

Part b)

[tex]F = 62.5 N[/tex]

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

[tex]Fcos\theta = mg + F_f[/tex]

[tex]Fsin\theta = F_n[/tex]

so we have

[tex]Fcos\theta = mg + \mu(Fsin\theta)[/tex]

[tex]F(cos\theta - \mu sin\theta) = mg[/tex]

[tex]F = \frac{mg}{cos\theta - \mu sin\theta}[/tex]

[tex]F = \frac{55}{cos50 - 0.310(sin50)}[/tex]

[tex]F = 135.7 N[/tex]

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

[tex]Fcos\theta = mg - F_f[/tex]

[tex]Fsin\theta = F_n[/tex]

so we have

[tex]Fcos\theta = mg - \mu(Fsin\theta)[/tex]

[tex]F(cos\theta + \mu sin\theta) = mg[/tex]

[tex]F = \frac{mg}{cos\theta + \mu sin\theta}[/tex]

[tex]F = \frac{55}{cos50 + 0.310(sin50)}[/tex]

[tex]F = 62.5 N[/tex]

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