Bob is pulling a 30 kg filing cabinet with a force of 200 n, but the filing cabinet refuses to move. the coefficient of static friction between the filing cabinet and the floor is 0.80. what is the magnitude of the friction force on the filing cabinet?

Refer to the diagram shown.

F = 200 N, the applied force.

m = 30 kg. the mass of the cabinet.

W = mg = (30 kg) * (9.8 m/s²) = 294 N, te weight of the cabinet.

N = W, the normal reaction of the floor on the cabinet.

R = resistive frictional force.

μ = 0.80, the static coefficient of friction.

Because the cabinet does not move,

R = μN = 0.80 * (294 N) = 235.2 N

Because F < R, the cabinet will not move.

Answer: 235.2 N

The force of friction is equal to the normal force multiplied by the coefficient of friction. In this case, the normal force is the force of gravity. F=ma, where m is the mass of the object and a is acceleration. The acceleration due to gravity is 9.8 m/s^2. Therefore F = 30kg * 9.8m/s^2 = 294n. (1n = 1kgm/s^2). friction force = normal force * coefficient of friction friction force = 295n * 0.80 = 235.2n