Question

If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move the crate at a steady speed across the floor

Answer 1

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

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Vertically the forces acting on the object:

• Normal force $N$ (acting up from the object)

• weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$, so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$, and:

$0=N-mg\\N=mg$

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now let's analyze the horizontal forces

• frictional force: $f= \mu N$ and since $N=mg$  --> $f=\mu mg$

• force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

$F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N$