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

58.27 N

Step-by-step 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`