Question

A mass of 5 kg is placed at rest on a horizontal surface with a coefficient of static friction of 1.03 and a coefficient of kinetic friction of 0.68. a force is of 26 newtons is applied to it downward in the -y direction and another force of 26 newtons is applied horizontally in the +x direction. what is the amount of friction, in newtons, acting on the mass? note: this could be static or kinetic friction... you must determine which kind of friction is acting here

Answer 1

Until the object starts to move, the static coefficient of friction must be used since the object is at rest initially.  If the horizontal force is large enough to overcome the friction force and the object accelerates, we switch to the kinetic coefficient.
The total downward force in the -y direction is the weight plus the applied 26N:
Fy=mg-26N = 5*-9.81-26 = -75.05N
The maximum friction force here is the magnitude of this force times the static coefficient of friction.  This force will point in the opposite direction as an applied horizontal force, no matter which way it points (non conservative forces tend to oppose motion in all directions)
Ff=75.05N*1.03=77.3N
Note this is greater than the applied horizontal x-directed force of 26N.  This means the answer is 26N of friction force acting on the object.  Why not the full 77.3N?  Because that is the maximum amount you can oppose along the x-direction before the object starts to move.  But since we are only applying 26N, that is all the friction force pushes back with. This is a statement of Newton's 3rd Law of equal and opposite actions.  If the object pushed back with the full 77.3N in this case it would accelerate backward (F=ma), since there would be a net force on the object (77.3N - 26N = 51.3N)  This obviously doesn't happen!