Answer to Question #180487 in Mechanics | Relativity for Tim Jefferson

Question

Peter wants to move a cabinet, with a mass of 90 kg ,across the floor.

1.what is the magnitude of static friction force before Peter touches the cabinet?

2.He increases the force to 200N , but tge cabinet does not move. what is the static frictional force?

3. A force of 350N is needed to bring the cabinet into motion. calculate the coefficient of static friction. 

4. when Peter continues to push the crate ,he exerts a force of 300N to keep it moving at a constant velocity. calculate the coefficient of kinetic friction.​

Answer 1

1.) For an object at rest on a flat surface, static friction is zero

So, the value of static friction is zero before Peter touches the cabinet.

2.)  If you push horizontally with a small force, static friction establishes an equal and opposite force keeps the cabinet at rest.

So, If Peter  increases the force to 200N , but the cabinet does not move then the value of static friction is equal to the force applied by Peter which is equal to 200N

3.) As Peter push harder, the static friction force increases to match the force. Eventually maximum static friction force is exceeded and the cabinet just moves.

So,

$$\mu_s=\dfrac{f_{s(max)}}{N}=\dfrac{350}{90\times 10}=0.39$$

4.) And as Peter continues to push the crate with 300N force,

So the value of coefficient of kinetic friction $\mu_k=\dfrac{\text{Applied Force}}{N}=\dfrac{300}{900}=0.33$