Question

A 5.0-kg crate is on an incline that makes an
angle of 30° with the horizontal. If the coeffi-
cient of static friction is 0.50, the minimum force
that can be applied parallel to the plane to hold
the crate at rest is:
(A) O
(B) 3.3N
(C)30N
(D) 46N​

Answer 1

Hi,

Answer:

The minimum force  that can be applied parallel to the plane to hold  the crate at rest is 3.3 N.

Explanation:

Given data:

Mass of the crate, m = 5.0 kg

The crate is inclined at an angle θ = 30° with the horizontal

Coefficient of friction, µ = 0.50

To find: Minimum force applied to the plane to hold the crate at rest

Let the acceleration due to gravity, g = 9.8 m/s²

We know the there is a force acting on the body due to the weight in the downward direction whose horizontal component is “mg sinθ” and vertical component is “mg cosθ”. Another force “N” will be acting just normal to the plane and a frictional force “f” is also acting just opposite to the mg sinθ up the plane as shown in the figure below.

Therefore,

Force acting down the plane due to the weight,  

= mg sinθ

= 5 * 9.8 * sin 30°  

= 24.5 N  

The frictional force acting up the plane,  

f = µ * N = µmg cos θ = 0.5*5*9.8*cos 30° = 21.217 N

Thus,  

The minimum force acting on the plane to hold the crate at rest = 24.5 – 21.217 = 3.283 N = 3.3 N

Hope it helps!!!!