The block in the figure below remains motionless against the wall because of an applied force

Question

The block in the figure below remains motionless against the wall because of an applied force and the force of static friction between the block and the wall. The coefficient of static friction is μs. Draw a FBD of the block. Determine the force of friction in terms of the variables described in the problem.

Answer 1

Without being given the geometry of the problem - you have not provided the picture, a definitive answer can't be provided.

in ANY problem with forces, especially if friction is included, you will need to solve in both the x and y-directions. The gravitational force, F_g, will always be present and always act down.

The normal force, N, always acts perpendicular to the surface and exactly balances all other forces or components of forces that area perpendicular to the surface. Friction forces, F_f, always oppose motion or the forces that would otherwise cause motion. The magnitude of the static friction is any value from 0 to u_sN. where u_s is the coefficient of static friction and N is the normal force. Kinetic friction is u_kN. where u_k is the coefficient of static friction and N is the normal force. Kinetic friction does not change unless normal force changes.

In this case, assume the wall is vertical (if it's not, then the gravitational force will need to be broken down into components). Define the positive x-direction the direction of the normal force and the positive of the y-direction up. The normal force will act perpendicular to the wall. Gravitational force will be parallel and in the negative (down) direction. The direction and magnitude of the frictional force will depend on the direction and magnitude of the applied force. The applied force, F_A, has an x-component in the opposite direction of the normal force and is equal and opposite to the normal. F_Ax= -F_A cos (Θ), where Θ is the angle between the force and the negative x-axis. the " - " is because the force is in the negative direction as defined above. F_Ay= F_A sin (Θ); this may be negative or positive depending on the angle - if the angle is below the x-axis, it is negative, above it is positive; if the applied force is perpendicular to the wall, then the y-component is 0.

Σ F_x = 0 = N - F_A cos (Θ) ==> N = F_A cos (Θ)

Σ F_y = 0 = F_A sin (Θ) + μN - mg ⇒ μN = mg - F_A sin (Θ)

Important: you should have the value μN ≤ u_sN. If this is not true, then the box is moving, and the equation for y becomes: Σ F_y = ma = F_A sin (Θ) + μ_kN - mg ⇒ μ_kN = ma + mg - F_A sin (Θ)

NOTE: if you define your coordinate system and/or angle differently, your equations will look different, but the final answer (magnitude and direction) will be the same..