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

Answer:

$\mu > 0.225$

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:

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

$\mu > \frac{4,500N}{20,000N}$

$\boxed{\mu > 0.225.}$