A 1,952-kg car is moving down a road with a slope (grade) of 10% while slowing d

Question

A 1,952-kg car is moving down a road with a slope (grade) of 10% while slowing down at a rate of 3.2 m/s^2. What is the direction and magnitude of the frictional force?

AND

A 1,834-kg car is moving down a road with a slope (grade) of 11% while speeding up at a rate of 3.5 m/s^2. What is the direction and magnitude of the frictional force?

When setting up Newton's 2nd Law, you always first write the sum over all components, e.g.,

If you know everything but fx, you can solve for fx. Your solution will give you a sign, "+" or "-", and the magnitude of the friction.

Once you finished the set, look back and try to understand what is happening in each situation. In some cases, the friction might have turned out to be pointing opposite to your intuition. Why is that? For each situation, think about what you as the driver do to achieve the stated acceleration.

Explanation / Answer

Frictional force Ff = umg cos theta

here theta = friction = tan theta = 0.1

theta 5.71 deg

accleration a = g sin theta

a = 9.81 sin 5.71

a = 0.976 m/s^2

frictional force FF = ma = 1952 * (3.2-0.976)

Ff = 4341.248 N

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AND

tan theta = 0.11

theta = 6.277 deg

a = 9.81 sin 6.277

a = 1.072 m/s^2

Ff = 1834 *(9.81 - 1.072)

Ff = 16025.5 N

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