A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Figure 1)). The floor is rough enough to prevent the crate from slipping

Question

Alvin

Physics

10 September, 01:01

A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Figure 1)). The floor is rough enough to prevent the crate from slipping

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Cherish Lucero · Answer 1

This problem has three questions I believe:

> How hard does the floor push on the crate?

We have to find the net vertical (normal) Fn force which results from Fp and Fg.

We know that the normal component of Fg is just Fg, which is equal to as 1110N.

From the geometry, the normal component of Fp can be calculated:

Fpn = Fp * cos (θp)

= 1016.31 N * cos (53)

= 611.63 N

The total normal force Fn then is:

Fn = Fg + Fpn

= 1110 + 611.63

= 1721.63 N

> Find the friction force on the crate

We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely, so we can say that the horizontal component is zero:

Fh = Fph + Fgh

= (Fp * sin (θp)) + 0

= 1016.31 N * sin (53)

= 811.66 N

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

μs = Fh / Fn

= 811.66 N / 1721.63 N

= 0.47