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Fictional force problem given a slope(grade)?

  • #1
fictional force problem....given a slope(grade)???

A 1000kg car is moving down a road with a slope(grade) of 15% at a constant speed of 15m/s. What is the direction and magnitude of the frictional force?

So...

V= 15m/s
a= 0 (constant)

The slope really throws me off..I don't know where to start! And the percent.... What am I supposed to do with that??
 
Last edited:

Answers and Replies

  • #2
1,198
5


Look up what grade means. If there were no friction, the car would be accelerating. But it isn't. Therefore what can you say the friction force must equal?
 
  • #3


Zero? The frictional force would be zero. So, then the grade (the incline) was given just to make us think and try to picture it, I guess.

Ok, how about if we're given the grade and told that the the car was speeding up at 3m/s^2

I know the mass (1000kg) and I think theta would be the percentage of 15? Would the velocity be the force N or friction f? and where would I plug in my acceleration?

I think f=ma ; Ncos(theta) + fcos(theta)-mg

I hope I'm on the right track...the prof. has only touched on the subject and he hasnt assigned a textbook..so really I'm trying to google what I can but its more complicated than I can do on my own. Any help would be appreciated :)
 
  • #4
1,198
5


Friction is not zero. If it were zero, the car would be accelerating. It is not. It is moving at constant speed.

For the car to move DOWN the slope at a CONSTANT speed, what forces must be equal and opposite.
 
  • #5


Would it be equal to the mass?
 
  • #6
1,198
5


Mass is not force. Mass times acceleration is force. The force that propels the car down the slope is a function of mass, gravity (acceleration), and a trigonometric function of the angle of the slope.

You should draw a free body diagram of the vehicle on the slope and determine what the acting forces are. Obviously, gravity and the downhill directions do not coincide. Therefore there is a trigonometric relationship between them.
 

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