MCAT Physics Module I: Friction

T/F: Friction is always an opposing force to the x-component of gravity

F; Friction does not always oppose the x-component of gravity. Friction is a force that opposes the relative motion or tendency of motion between two surfaces in contact. While it often acts opposite to a component of gravity when something is sliding down an incline, it could act in the same direction as the x-component of gravity if an external force is trying to push an object up an incline, or if it's static friction preventing motion up an incline.

T/F: Static friction is mostly lower than than kinetic friction.

Kinetic friction can be lower than static friction [ μ_k<_ μ_s]

Derive the coefficient for static friction:

magnitude of the force to budge the object/ magnitude of normal force

Derive the coefficient of kinetic friction:

the magnitude of the force of friction/ magnitude of normal force otherwise known as tan(theta).

T/F: Friction is dependent upon surface area.

False; friction is independent of surface area and only dependent on the type of surface

How can you find the magnitude of the FORCE of the static friction? What about kinetic friction?

Fstatic <_ μ_s*F_N; Fkinetic= μ_k*F_N

Suppose an object is resting on a surface. If we apply a force less than the max force static friction, the object will not move. As soon as the applied force exceeds the max static force, the object will begin to slide and kinetic friction will replace static friction. Let F_n= 100 N, μ_s= 0.7, μ_k=0.3.

Fstatic= (0.7) 100N= 70 N; Fkinetic= 0.3 (100 N)= 30 N

Friction is a _____ force, its _______ and _______ can change.

resistive; direction and magnitude

Newton can also be written as …

kg*m/s²

the magnitude of the gravitational force between two objects is ….

Fg= Gm₁m₂/r²

Find the gravitational force between an electron and a proton that are 10^-11 m apart. (*Note: mass proton=1.67×10^-27; mass of electron= 9.11×10^-31 kg)

Using Newton’s Law of gravitation:
Fg=(6.67×10^-11 Nm²/kg²)(1.67×10^-27 kg)(9.11×10^-31 kg)/(10^-11 m)²= 10^-45

T/F: Just because a object is stationary does not mean it is experiencing a maximal amount of friction.

True.

A wheel is rolling along a road. Does it experience kinetic firction?

No, becuase the tire is not actually sliding along the pavement. It maintains an instaneous point of contact with road and therefore static friction.

When calculating fricitonal forces, how is it directionally assigned?

the direction of the frictional force always opposes the movement. the frictional force must be in the opposite direction of the velocity vector.