Find frictional force given acceleration, grade, and mass

  • #1
Butterfly30
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a 1000 kg car is moving down a road with slope (grade) of 15% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)


Using f=ma

I have the mass 1000 kg, the acceleration 3.7m/s^2, and slope 15%

Would my force be the tangent of the angle 15??
 

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  • #2
SammyS
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a 1000 kg car is moving down a road with slope (grade) of 15% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)

Using f=ma

I have the mass 1000 kg, the acceleration 3.7m/s^2, and slope 15%

Would my force be the tangent of the angle 15??
Without working the problem, I say no. For one thing the units are incorrect.
 
  • #3
Butterfly30
28
0
Sorry these were the numbers, I accidently mixed two probs together:/


a 1892kg car is moving down a road with slope (grade) of 12% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)


Using f=ma

I have the mass 1000 kg, the acceleration 3.7m/s^2, and slope 15%

Would my force be the tangent of the angle 15??
 
  • #4
Butterfly30
28
0
So basically given this problem:


a 1892kg car is moving down a road with slope (grade) of 12% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)

Would my force be the tangent of the angle 12??
With f=ma
 
Last edited:
  • #5
SammyS
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Sorry these were the numbers, I accidentally mixed two probs together:/

a 1892kg car is moving down a road with slope (grade) of 12% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)

Using f=ma

I have the mass 1000 kg, the acceleration 3.7m/s^2, and slope 15%

Would my force be the tangent of the angle [STRIKE]15[/STRIKE] 12??

The net force will be in a direction parallel to the ramp (down the ramp).

If the ramp makes an angle of θ with respect to the horizontal, then tan(θ) = 0.12 .

Draw a free body diagram.
 

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