Question

A cable capable of pulling 4,500 N snapped while trying to drag a 20,000 N compressor across the street. What is the coefficient of static friction for this scenario?

Note: the 4500 N cable is used as Fs and is calculated the same as Fk
0.225
> 0.225
4,500 N
> 4,500 N
4.44
> 4.44
Please explain to me how you get your answer.

Answer 1

>0.225 is correct if its not please let me know I'm willing to help!

Answer 2

`\mu > 0.225`

Step-by-step explanation:

The cable snapped because the frictional force
`f_s` was greater than the tension in the rope:

`f_s > 4,500N`

Now,

`f_s =\mu N`

where
`\mu` is the coefficient of static friction, and
`N` is the normal force. In our case, the normal force on the compressor is 20,000 N; therefore,

`f_s =\mu (20,000N )`.

Putting this into the condition
`f_s > 4,500N`, we get:

`\mu (20,000N )> 4,500N`

`$\mu > (4,500N)/(20,000N) $`

`\boxed{\mu > 0.225.}`