Question

A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate

moving. What is the coefficient of static friction between the crate and the floor?

Answer 1

0.456033049

Step-by-step explanation:

`F=\mu N` where N=mg hence

`F=\mu mg` where m is mass of object, g is acceleration due to gravity whose value is taken as
`9.81 m/s^(2)`,
`\mu` is the coefficient of static friction and F is the applied force.

Making
`\mu` the subject we obtain

`\mu=\frac {mg}{N}` and substituting m for 38 Kg, g for
`9.81 m/s^(2)` and 170 N for F we obtain

`\mu=\frac{170} {38*9.81}=0.456033049`

Therefore, the coefficient of static friction is 0.456033049