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Acceleration of car on a gradient

  • Thread starter Molly1235
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  • #1
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"What is the acceleration of a car of mass 1200kg, when the driving force on it is 4000N and the total frictional force on it is 900N?

What is the acceleration of this car climbing a hill with a gradient of 8 degrees?"

I got the first part, as obviously a= f/m, resultant force is 3100, and 3100/1200 = 2.6 ms^-2.

But I'm really not sure how to approach the second part...can anyone give me some clues??

Thanks!
Molly :-)
 

Answers and Replies

  • #2
PhanthomJay
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When the car is on the hill that makes an 8 degree angle with the horizontal, now both gravity and friction is working against the driving force, so its acceleration will be less. What is the component of the gravity force along the incline? What is the net force acting on the car?
 
  • #3
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I found the vertical component to be 125N??
 
Last edited:
  • #4
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There is the normal push from the inclined plane, which will be different than for horizontal plane.
So the friction will be slightly different too.
The largest effect comes from the tangential (along the ramp) component of the weight.
 
  • #5
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There is the normal push from the inclined plane, which will be different than for horizontal plane.
So the friction will be slightly different too.
The largest effect comes from the tangential (along the ramp) component of the weight.
Ok that really confused me...I'm sorry!! :(
 
  • #6
PhanthomJay
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Forget about force triangles. Choose the x axis as the axis parallel to incline, and the y axis as the axis perpendicular to the incline. Assume friction force is 900 N. Draw free body diagram of car, identify all forces acting, and apply F_net = ma in x direction. Be sure to correctly determine componets of the weight in the chosen x and y directions. See here for a brief tutorial:
http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html
 
  • #7
gneill
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There is the normal push from the inclined plane, which will be different than for horizontal plane.
So the friction will be slightly different too.
The largest effect comes from the tangential (along the ramp) component of the weight.
I think you can safely ignore any change in friction; the given friction value must pertain to losses in the drive train and perhaps air resistance. The given value for friction is certainly not enough to transmit the driving force to the road, so some unspecified static friction (between tires and road) is responsible for that and one can assume it is sufficient for all cases.
 
  • #8
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You are making it too complicated. This is not a real situation analysis but a school textbook problem. I agree that it is not clear what kind of friction is the one given and it does not matter for the first part of the problem.
Anyway, even assuming that it is some sort of kinetic friction, you can ignore the change on the ramp because it's second order in the angle whereas the tangential component of the weight is first order.
 
  • #9
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I've got it now!! Got the acceleration as 1.2 ms^-2 which matches the book - thanks guys! :-)
 

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