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?

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

N = mg

N = 40 × 9.8

Normal reaction (N) = 392 N

Coefficient of friction (μ) = 0.69

F = μN

F = 0.69 × 392

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

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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)

i don't understand. explain it clearly

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