Question

The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is µs =0.25.

(A)Determine the minimum horizontal force P required to hold the crate from sliding down the plane.
(B) Determine the minimum force P required to push the crate up the plane

Answer 1

To determine the minimum horizontal force required to hold the crate from sliding down the plane, calculate the maximum force of static friction. To determine the minimum force required to push the crate up the plane, calculate the maximum force of kinetic friction.

Step-by-step explanation:

To determine the minimum horizontal force required to hold the crate from sliding down the plane, we need to consider the maximum force of static friction. The maximum force of static friction can be calculated by multiplying the coefficient of static friction (µs) by the normal force (N). In this case, the normal force is equal to the weight of the crate (mass × gravitational acceleration). Therefore, the maximum force of static friction is µs × weight. The minimum horizontal force required to hold the crate from sliding down the plane is equal to the maximum force of static friction.

To determine the minimum force required to push the crate up the plane, we need to consider the maximum force of kinetic friction. The maximum force of kinetic friction can be calculated by multiplying the coefficient of kinetic friction (µk) by the normal force (N). In this case, the normal force is equal to the weight of the crate (mass × gravitational acceleration). Therefore, the maximum force of kinetic friction is µk × weight. The minimum force required to push the crate up the plane is equal to the maximum force of kinetic friction.

Answer 2

Step-by-step explanation:

Given

mass of crate
`m=50 kg`

Coefficient of static Friction
`\mu =0.25`

Let
`\theta`be angle of inclination

resolving Forces along and perpendicular to plane

Force along Plane

`P\cos \theta -mg\sin \theta -f_r=0`----------1

Forces perpendicular to plane

`P\sin \theta +mg\cos \theta -N=0`

`f_r=\mu N`

`f_r=\mu (P\sin \theta +mg\cos \theta)`

Substitute the value of
`f_r`

`P\cos \theta -mg\sin \theta -\mu (P\sin \theta +mg\cos \theta)=0`

`P\left [ \cos \theta -\mu \sin \theta \right ]=mg\left [ \sin \theta +\mu \cos \theta \right ]`

`P=mg\cdot (\sin \theta +\mu \cos \theta )/(\cos \theta -\mu sin \theta )`

(b)Minimum Force require to push the crate up the Plane

i.e. Force Must act parallel to plane

Forces Along the Plane

`P-mg\sin \theta -f_r=0`-------2

Forces Perpendicular to plane

`mg\cos \theta -N=0`

`f_r=\mu N`

`f_r=\mu mg\cos \theta`

Substitute in Equation 2

`P=mg\sin \theta +f_r`

`P=mg(\sin \theta +\mu \cos \theta )`