Question

Two workers are sliding 390 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 200 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Answer 1

The coefficient of kinetic friction
`\mu= 0.16989`

Step-by-step explanation:

From Newton's second law

`\sum\overset{\rightarrow}{F}=m\cdot\overset{\rightarrow}{a}`

If the velocity is constant, that means the summation of all forces must be equal to zero. Draw the free-body diagram to obtain the sums of forces in x and y. It must include the Friction Force, in the opposite direction of the displacement, the weight (
`W=mg=390*9.81=3825.9N`), the Normal Force, which is the is the consequence of Newton's third law and the forces from the two workers.

The sum in y is:

`\sum F_(y)=F_(N)-3825.9=0`

Solving for the
`F_(N)`:

`F_(N)=$ $3825.\,\allowbreak9N`

The sum in x is:

`\sum F_(x)=450+200-F_(f)=0`

Solving for the
`F_(f)`:

`$F_(f)=650.0N`

The formula of the magnitude of the Friction force is

`F_(f)=\mu F_(N)`

That means the coefficient of friction is:

`\mu=(F_(f))/(F_(N))=(650.0)/(3825.\,\allowbreak9)=\allowbreak0.16989`