Question

An object with a mass of `5 kg` is on a ramp at an incline of `pi/8`. If the object is being pushed up the ramp with a force of `2 N`, what is the minimum coefficient of static friction needed for the object to remain put?

Answer 1

`color(blue)(0.37)color(white)(8888)` 2 d.p.

Explanation:

Since the `bb(2 N)` force is acting upwards the frictional force `bb(Fr)` is acting in the opposite direction.

For the body to remain static, we have:

`Fr=muR`

Where `mu` is the coefficient of friction.

Resolving forces perpendicular to the plane:

`R=5gcos(pi/8)`

Resolving forces horizontal to the plane:

`Fr=5gsin(pi/8)-2N`

Substituting these two equations into`\ \ \ \ \Fr=muR`:

`5gsin(pi/8)-2=mu*5gcos(pi/8)`

`:.`

`mu=(5gsin(pi/8)-2)/(5gcos(pi/8))`

Taking `g=9.81`

`mu=(5*(9.81)sin(pi/8)-2)/(5*(9.81)cos(pi/8))=0.3700793240N`

`mu=0.37` 2 d.p.