- #1

Butterfly30

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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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- Thread starter Butterfly30
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- #1

Butterfly30

- 28

- 0

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??

- #2

SammyS

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Without working the problem, I say

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??

- #3

Butterfly30

- 28

- 0

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

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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