Question

A 40-kilogram crate is at rest on a level surface. If the coefficient of static friction between the crate and the surface is 0.69, what horizontal force is required to get the crate moving?

Answer 1

Answer:

270.8 N

Explanation:

y axis :

N-w=0

N=mg = 40*9.81 =392.4 N

x axis :

P - fs =0

p = meo * N = 0.69 * 392.4 = 270.8 N (ans)

Answer 2

The horizontal force required to get the crate moving is 270.48 N

We'll begin by calculating the normal reaction. This can be obtained as follow:

Mass (m) = 40 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (N) =?

N = mg

N = 40 × 9.8

N = 392 N

• Finally, we shall determine the force required to move the crate.

Normal reaction (N) = 392 N

Coefficient of friction (μ) = 0.69

Force (F) =?

F = μN

F = 0.69 × 392

F = 270.48 N

Thus, the force needed to move the crate is 270.48 N

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