# Finding force parallel, HELP

1. Dec 1, 2008

### vballkatie22

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

A 50kg crate rests on the floor. The coefficient of static friction is .5. The force parallel to the floor needed to move the crates is most nearly:

Choices:
50N
125N
250N
500N

2. Relevant equations
I have no idea what equation to use for this equation.

3. The attempt at a solution

No attempts yet since I don't know which equation to use.

2. Dec 1, 2008

### rock.freak667

Do you know the equation that F=uR, where F is the frictional force and R is the normal reaction?

3. Dec 1, 2008

### vballkatie22

I have seen it before but my teacher never really used it. But I would like to learn how to use it. Can you show me?

4. Dec 1, 2008

### rock.freak667

Firstly, you'll need to find the normal reaction,R. In this case it is simply the crate's weight. When you have that, put that into F=uR where u is the coefficient of friction. Then compare the value of friction that you get to the answers given.

5. Dec 1, 2008

### vballkatie22

Ok. That sounds pretty easy. So for that it would look like this..

F= uR
F= (.5)(50)
I guess I multiply right? And that gives me 25N but that isn't an answer.

Did I do something wrong?

6. Dec 1, 2008

### rock.freak667

50kg is the crate's mass. To find its weight you need to multiply the mass and the acceleration due to gravity

7. Dec 1, 2008

### vballkatie22

Ok. So to find the weight I take 50kg and times that by acceleration which is 9.8 (gravity)
To get 490.

Correct?

8. Dec 1, 2008

### rock.freak667

yes that is the weight of the crate and hence its normal reaction. Now what is the frictional force (this also is the frictional force which causes the crate to be in limiting equilibrium)

9. Dec 1, 2008

### vballkatie22

Would it be .5? I don't really know.

10. Dec 1, 2008

### rock.freak667

Try F=uR => F=(0.5)(490)

11. Dec 1, 2008

### vballkatie22

Ok.. so the answer would be 245N but the closest one I have is 250N.

Thanks!!