physics - friction

[msjl33]

These quetions come from Dr. Biehle's "The MCAT PHysics Book" (Nova Press); chapter 6

1. A car (mass m) is going up a shallow slope (angle x with the horizontal) when the driver sees a red light and suddenly applies the brakes. The car goes into a skid as it comes to a stop. The static coefficient of friciton between the tires and the road is MUs, and the kinetic coefficient of friction is MUk. (note: MU represents the greek letter u)

Which expression gives the force of friction on the car?
a) mg
b) mg sin x
c) MUk*N
d) MUs*N

answer (C). But why is it not d? If the car is skidding to a STOP, aren't we dealing with static friction, rather than kinetic friction?


2. A man is trying to push a washer (100kg) along a level floor, but the washer is not moving. He is pushing a horizontal force 700N. The coefficients of friciton are MUs = 0.8 and MUk = 0.6.

How hard would the man have to push to get the washer moving?
a) 600N
b) 700N
c) 800N
d) 1000N

answer (C).
I thought it was D. I thought that whatever the force the man exerts, it has to overcome the static frictional force, which would be = 0.8N = 0.8(1000) = 800N. So, how is 800N enough to overcome the static force of 800N?



I would really appreciate your time and patience to explain either or both of these questions. Thanks!!

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[Hamiltonian]

These quetions come from Dr. Biehle's "The MCAT PHysics Book" (Nova Press); chapter 6

1. A car (mass m) is going up a shallow slope (angle x with the horizontal) when the driver sees a red light and suddenly applies the brakes. The car goes into a skid as it comes to a stop. The static coefficient of friciton between the tires and the road is MUs, and the kinetic coefficient of friction is MUk. (note: MU represents the greek letter u)

Which expression gives the force of friction on the car?
a) mg
b) mg sin x
c) MUk*N
d) MUs*N

answer (C). But why is it not d? If the car is skidding to a STOP, aren't we dealing with static friction, rather than kinetic friction?


2. A man is trying to push a washer (100kg) along a level floor, but the washer is not moving. He is pushing a horizontal force 700N. The coefficients of friciton are MUs = 0.8 and MUk = 0.6.

How hard would the man have to push to get the washer moving?
a) 600N
b) 700N
c) 800N
d) 1000N

answer (C).
I thought it was D. I thought that whatever the force the man exerts, it has to overcome the static frictional force, which would be = 0.8N = 0.8(1000) = 800N. So, how is 800N enough to overcome the static force of 800N?



I would really appreciate your time and patience to explain either or both of these questions. Thanks!!

With the first question, you just have to visualize that the car is experiencing the friction while it's moving.

 

[bxadook]

The coefficient of kinetic friction is used whenever the problem states that the particle or object is currently in motion. The coefficient of static friction is used whenever the object is not in motion (hence static).

Part two I think is referring to the minimum amount of force required to put the washer into motion. The object weighs 100kg and has a coeff of static friction of 0.8. Therefore: (.8)(100kg)(9.81m/s^2) = 785N is referring to the amount of force required to overcome the static frictional force. I think it rounds to 800N. Remember, it is asking for the minimum amount of force. 1000N would surely put the washer into motion but that is not what the question was asking.

 

[fizzgig]

note for the 1st if the car were rolling WITHOUT skidding to a stop, the static coefficient is still the one to use. with skidding the kinetic coefficient is in play until the car actually comes to a complete halt.

for the 2nd i think as soon as your applied force > mu_s*F you get slipping, and right then and there kinetic friction takes over, which is why pushing gets easier once it's moving.

EDIT my bad, looks like i was wrong, fixed it, above dude correct and i guess it's just bc of rounding.
this is why i look through these forums

 

[FrkyBgStok]

with friction i always imagine a block of mass on ice. if the block is moving it is kinetic, and in this case, your wheels are not moving, but there is a slide, so kinetic.

agree with other answers as well.

 

[puravida85]

First one, I agree with everyone else.. The car is skidding so it's moving, therefore kinetic friction.

Second one, I think the second question is a pain in the butt because as someone stated above, the Static Friction's force is a max of 785N. Therefore anything above that, you'll get the box object moving. of course we're so use to rounding, we assume that the max static friction force is:

F = Mus * Nf

Nf = 100 * 10 , but in reality it should be 100 * 9.8.

So with 800 N of force, that thing is going! Now you can also find the acceleration a = F /m. Anyways, just be more careful next time.