Question

An object with a mass of `8 kg` is on a ramp at an incline of `pi/8`. If the object is being pushed up the ramp with a force of `4 N`, what is the minimum coefficient of static friction needed for the object to remain put?

Answer 1

`0.36`

Explanation:

Here downward component of the weight of the object tends to pull it down along the ramp.

It's value is `mg sin ((pi)/8)=30N`

Now,given externally `4N` force is applied upward along the plane,so net force on the object becomes`30-4=26N` acting downwards along the ramp.

So,static frictional force must act by that amount upwards,such that the net force acting on the object will be zero.

Now,here maximum value of frictional force that can act is `mumg cos ((pi/8))=72.42 mu`(where, `mu` is the coefficient of static frictioanl force)

So,we can write,

`72.42 mu=26`

or, `mu=0.36`