# Find frictional force given acceleration, grade, and mass

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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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.

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??

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:
SammyS
Staff Emeritus
Homework Helper
Gold Member
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.