Friction and Constant velocity

  • Thread starter CaptainSFS
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Homework Statement


A car of mas m = 1000kg is sliding down a hill. The coefficients of friction between the car's tires and the ground are us=0.89 and uk=0.61. For what inclination angle will the car slide down the hill with a constant velocity?


Homework Equations





The Attempt at a Solution



Now I realize that because it is constant velocity, the acceleration is 0, so the summation of the forces needs to be 0, but i don't know how to set this up. Any help?

Thanks. :)
 

Answers and Replies

  • #2
tiny-tim
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Now I realize that because it is constant velocity, the acceleration is 0, so the summation of the forces needs to be 0, but i don't know how to set this up.

Hi CaptainSFS! :smile:

Hint: there are only three forces … the weight, the normal force, and the friction force … so set up a force triangle :smile:
 
  • #3
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hello, yes, so i have my two frictional forces pointing upwards on the slope. I set it up like this... (1000)(9.81cos(theta))(.89)+(1000)(9.81cos(theta))(.61)-?=0 ...If this is correct, which I am not so sure b/c it just seems like I wouldn't be able to create an opposite force to equal it, so that the acceleration is 0. These frictional forces really confuse me, and I feel like I keep setting it up wrong.
 
  • #4
tiny-tim
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hello, yes, so i have my two frictional forces pointing upwards on the slope. I set it up like this... (1000)(9.81cos(theta))(.89)+(1000)(9.81cos(theta))(.61)-?=0 ...If this is correct, which I am not so sure b/c it just seems like I wouldn't be able to create an opposite force to equal it, so that the acceleration is 0. These frictional forces really confuse me, and I feel like I keep setting it up wrong.

Nooo …

i] you can't have both kinetic friction and static friction … it's one or the other

ii] what about the weight force?
 
  • #5
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okay, so I'm assuming them I will be using the static coefficient because that's the kind tires use on pavement and whatnot. For the weight force... would that be the force pulling the car down the slope? m*g*sin(theta)?

or do u mean the weight force as 1000*9.81?
 
  • #6
tiny-tim
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okay, so I'm assuming them I will be using the static coefficient because that's the kind tires use on pavement and whatnot.

Whoa!

that would be for wheels that are rolling

the car is sliding (in other words, the wheels are locked)
For the weight force... would that be the force pulling the car down the slope? m*g*sin(theta)?

or do u mean the weight force as 1000*9.81?

the weight force is mg, and you use its component in the required direction.
 
  • #7
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okay, so if I'm using it in the required direction, don't I have to use m*g*sin(theta) because gravity isn't pulling it from the side, it's pulling straight down on it?
 
  • #8
tiny-tim
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okay, so if I'm using it in the required direction, don't I have to use m*g*sin(theta) because gravity isn't pulling it from the side, it's pulling straight down on it?

Yes! Get on with it! :rolleyes:
 
  • #9
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thanks!
 

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