Question

A 5.0-kg crate is on an incline that makes an angle 30° with the horizontal. If the coefficient of static friction is 0.5, what is the maximum force that can be applied parallel to the plane without moving the crate?

Answer 1

Answer:

$F_{applied} = 45.8 N$

Explanation:

When maximum force is applied on the crate along the plane so that it will not move then in that case friction force and component of the weight of the crate is along the plane opposite to the applied force

So here we will have

$mgsin\theta + F_f = F_{applied}$

now we know that

$F_f = \mu F_n$

also we know that

$F_n = mg cos\theta$

so we will have

$F_f = \mu ( mg cos\theta)$

now we will have

$mg sin\theta + \mu (mg cos\theta) = F_{applied}$

$(5)(9.81)sin30 + (0.5)(5)(9.81)cos30 = F_{applied}$

so we will have

$F_{applied} = 45.8 N$