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Bob is pulling a 30kg filing cabinet with a force of 200N , 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?
Express your answer to two significant figures and include the appropriate units.

Respuesta :

MissPhiladelphia
Hi, thank you for posting your question here at Brainly.

First, let's enumerate the forces acting on the filing cabinet at rest. Forces along the y-direction are the weight (W) and the normal force (Fn). Since the object is at rest, Fn = W= mg = 30 kg*9.81 m/s^2 = 294.3

Forces along the x direction are opposite forces, the 200-N force and the friction force. The friction force is equal to 0.8*Fn = 0.8*294.3 = 235.44 N.

The 200-N force was not enough to overcome the friction force. Thus, Bob should apply at least 236 N for the cabinet to move.
carlosego

By making a free-body diagram in a vertical direction, the normal force is given by:

Fn = m * g

Where,

m: mass

g: acceleration of gravity.

Then, the friction force is given by:

F = μ * Fn

Where,

μ: friction coefficient

Substituting values we have:

F = μ * m * g

F = 0.80 * 30 * 9.8

F = 235 N

We note that the friction force (235 N) is greater than the 200 N force with which Bob pushes the sled.

Therefore, the sled will not move.

Answer:

The magnitude of the friction force on the filing cabinet is:

F = 235 N