A worker wants to turn over a uniform 1120-N rectangular crate by pulling at 53.

Question

A worker wants to turn over a uniform 1120-N rectangular crate by pulling at 53.0 ?(angle) on one of its vertical sides (the figure (Figure 1) ). The floor is rough enough to prevent the crate from slipping.

Part A

What pull is needed to just start the crate to tip?

Part B

How hard does the floor push on the crate?

Part C

Find the friction force on the crate.

Part D

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

PLS HALP! STEP BY STEP PLS!!

Explanation / Answer

length , Lx = 2.2 m

Ly = 1.5 m

let the pull be f

theta = 53 degree

w = 1120 N

A)

Torque around the lower left corner of the block

Ly *f*sin( theta) - Lx* W / 2 = 0

1.5 * f*sin(53) - 2.2*1120/2 = 0

f = 1028.42 N

the pull needed is 1028.42 N

B)

Forces on X and Y in center of mass

equating the forces vertically

N - F*cos(theta) - W = 0

N - 1028*cos(53) - 1120 = 0

N = 1738.92 N

the force exerted on the crate is 1738.92 N

C)

let the friction on the crate be Fr

F*sin(theta) - Fr = 0

Fr = 821.33 N

the friction force acting onn the crate is 821.33 N

(D)

The minimum static friction coefficient be us

Fr = us N

821.33 = us*1738.92

us = 0.47

the minimum coefficient of static friction needed to prevent the crate from slipping on the floor is 0.47

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