The magnitude of the friction force acting on the filing cabinet is 235.2 N.
The given in the problem are:
mass (m) = 30 kg
applied force (F) = 200 N
coefficient of static friction (μ)= 0.80
The unknown is the magnitude of the friction force.
The forces acting on the cabinet in the direction of motion (x- axis) are the applied force F and the friction force, f. Therefore,
[tex]F \ + \ (-f) \ = mg[/tex]
Static friction is the type of friction force acting on a stationary object. This is the force that must be overcome for the object to move.
The friction force,f, can be calculated by multiplying the coefficient of static friction by the normal force.
The normal force is the perpendicular force the surface applies on the object. If the surface is flat, the normal force is equal in magnitude but opposite in sign of the object's weight.
Mathematically, this is expressed as follows:
[tex]f = \mu mg\\f = \ (0.80)(30 \ kg)(9.8\frac{m}{s^2})[/tex]
[tex]f \ = 235.2 \ \frac{kg-m}{s^2}[/tex]
Since the object is not moving, as stated in the problem, then the applied force, F, should be less than the friction force,f.
The value calculated for f is 235.2 N which is less than the applied force of 200 N. This justifies the answer because indeed the friction force is greater than the force applied. The applied force is insufficient to overcome the friction force, hence, the object remains at rest.
Keywords: friction force, coefficient of static friction
