# What is the direction and magnitude of the frictional force?

## Homework Statement

A 2,200-kg car is moving down a road with a slope (grade) of 14% at a constant speed of 11 m/s.

## Homework Equations

The equation that I'm using is ƩF=ma and friction=μN

## The Attempt at a Solution

the mass is 2200kg
the acceleration is zero because of the constant speed.
I really cannot think of anyway to solve this solution.

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The frictional force always opposes the motion, it "eats up" the total energy in the system. To do any calculations with the frictional force we need at least the coefficient of friction μ.

And that the problem, mu is not given. Do I have to find mu in order to move on?

Well for this problem you do not need to find mu. Here are some steps to get you started:

1. Draw a free body diagram on the car.
2. Utilize Newtons Second Law in the x and y directions respectively
3. Algebraic Wizardry, manipulate the equations until you can find the frictional force using values you already know.

ok, but what about the 11m/s, does that velocity come into play?

ok, but what about the 11m/s, does that velocity come into play?
your question mentions that velocity is constant, and when velocity is constant the net force is always zero. so, here friction is equal to the component of gravitational force along the slope.

P.S. - like Legaldose already mentioned drawing free body diagram always clears out doubt.

ok so I solved the problem, and this is what I did: m*g*sin(7.96) the 7.96 is the reverse tangent.

so 2,200kg*9.81m/s^2*sin(7.96)= 2,989N

would that be considered my magnitude and direction?

rcgldr
Homework Helper
ok so I solved the problem, and this is what I did: m*g*sin(7.96) the 7.96 is the reverse tangent.
Not a big deal, but arctan(.14) ~= 7.97.

thanks, but what about the solution that I solved 2,989N is that solution the magnitude and direction, or if there is there still more?

haruspex