A 1.95 kg block is held in equilibrium on an incline of angle q = 53.1o by a horizontal force, F, applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is ms = 0.313, determine the minimum value of F.

Hmmm. What direction is the Force F, horizontal? or parallel to the plane?

horizontal

To determine the minimum value of the force F required to hold the block in equilibrium on the incline, we can consider the forces acting on the block.

First, let's break down the gravitational force acting on the block. We have the force of gravity (mg) acting vertically downward and it can be broken down into two components: the force pulling the block into the incline surface (mg * sin(q)) and the force perpendicular to the incline (mg * cos(q)), where q is the angle of the incline.

Since the block is in equilibrium, the force F must balance the force pulling the block into the incline surface (mg * sin(q)) and overcome the force of static friction (fs) between the block and the incline.

We can calculate the force of static friction using the formula fs = µs * N, where µs is the coefficient of static friction and N is the normal force.
The normal force N can be found as N = mg * cos(q).

So, fs = µs * N = µs * mg * cos(q).

To find the minimum value of F, we need to determine the maximum value of static friction that can be applied, which occurs just before motion is about to start. This is why we consider static friction instead of kinetic friction.

The maximum static friction force is calculated using the formula fs,max = µs * N.

Now, we can set up the equation for equilibrium in the horizontal direction:

F - fs,max = 0

F - µs * mg * cos(q) = 0

Solving for F, we have:

F = µs * mg * cos(q)

Substituting the known values:

F = (0.313) * (1.95 kg) * (9.8 m/s^2) * cos(53.1o)

Calculating this expression gives us the minimum value of F required to hold the block in equilibrium on the incline.