## Free Body Diagram

1. The problem statement, all variables and given/known data

A cavewoman is dragging her 65.4 kg mate home, up a gentle 10° incline. The co efficient of friction for skin on oil is 0.40. What is the minimum applied force she needs to keep the neanderthal moving with uniform motion?

2. Relevant equations

ff=μFn
fg=mg
fnet=ma

3. The attempt at a solution

I tried solving for Ff using ff=μFn but I don't have the Fn do I? We have gravity going down which is -9.8(65.4)=640.92...would that be the force of normal?
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 f=ma I have mass but not acceleration? How would i calculate it?
 so...fa=65.4(0)...?

## Free Body Diagram

it does not..thats my problem :S
 :/ Are you sure lol? I am pretty sure he prof wouldn't give a question thats not solveable
 ah alright, so prof musta made a mistake. Thanks, Got an exam tommorow though :P hope he doesnt put this question on it
 school in Calgary :P
 Recognitions: Gold Member The professor didn't make a mistake. You already calculated the force of gravity acting on the guy. For a slope of 10 degrees, what are the components of this force in the directions perpendicular and parallel to the incline? What direction does the force parallel to the incline point (up the incline, or down the incline)? If she is trying to pull the guy up the incline, what frictional force does she have to overcome? Make a list of the forces acting on the guy parallel to the incline. How many forces are there? According to the problem statement, is the guy accelerating up the incline, or is he just barely moving at a constant speed? Chet
 - Frictional Force, Moving towards the left - Force applied going [E 10 N] - Force of Gravity Down - Force of Normal Up Since it says gently, I assume it is moving at a constant velocity? or close to? What is the next step after Force of gravity? Thanks for replying Chet
 Recognitions: Gold Member The gravitational force acting on the guy can be resolved into two components, one perpendicular to the incline and the other parallel to the incline. What are the numerical values of these two components?
 component y = -9.7? Component x = Force of friction ? That is the only force acting in the x direction isn't it?
 I can't figure it out :/

Mentor
 Quote by ThePhysicsBoy component y = -9.7? Component x = Force of friction ? That is the only force acting in the x direction isn't it?
What do you mean by "the x direction"? If that is along the surface of the incline, then there is a force in addition to friction acting in the x direction, namely gravity. Gravity has a component along the incline.

Recognitions:
Gold Member
 Quote by ThePhysicsBoy component y = -9.7? Component x = Force of friction ? That is the only force acting in the x direction isn't it?
You already said that the gravitational force acting on the guy is 641 N. Do you know how to resolve this force into components perpendicular and parallel to the 10 degree incline?

Gravity doesn't exert any frictional force on the guy. The frictional force is exerted on the guy by the inclined plane. But we will get to that later.

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