Frictional force on a car on an inclined plane

  • #1
A 2319kg car is moving down a road with a slope grade of 11% and slowing down at a rate of 3.8m/s^2.Find the direction and magnitude of the frictional force ( define positive in the forward direction ie down the slope)

So the equation I have is f= ma+ mgsinθ. For my angle I get arctan(.11)= 6.27

Plug in. f= 2319*3.8 +2319*9.81*sin6.27 Ive already drawn a FBD. since the friction is moving forward it should be a negative answer (according to the question).... BUT I'm confused because if its slowing down is my acceleration supposed to be negative when I plug it in. This would make my answer negative....but would the fact that the acceleration is moving in negative direction cancel it out and make it positive?? I know I'm confusing myself I just don't know if my (a) should be negative and how it will effect the sign of the overall answer....help please
 

Answers and Replies

  • #2
CAF123
Gold Member
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A 2319kg car is moving down a road with a slope grade of 11% and slowing down at a rate of 3.8m/s^2.Find the direction and magnitude of the frictional force ( define positive in the forward direction ie down the slope)

So the equation I have is f= ma+ mgsinθ. For my angle I get arctan(.11)= 6.27

Plug in. f= 2319*3.8 +2319*9.81*sin6.27 Ive already drawn a FBD. since the friction is moving forward it should be a negative answer (according to the question).... BUT I'm confused because if its slowing down is my acceleration supposed to be negative when I plug it in. This would make my answer negative....but would the fact that the acceleration is moving in negative direction cancel it out and make it positive?? I know I'm confusing myself I just don't know if my (a) should be negative and how it will effect the sign of the overall answer....help please
Your relevant equation of motion is: [tex] mg\sin\theta - F_f = ma [/tex]
 
  • #3
Oh, but do u enter my (a) as negative since the car is slowing down?
 

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