# Physics friction

1. Feb 24, 2009

### Intrusionv2

1. The problem statement, all variables and given/known data
Find the normal force on an object, 80kg, which is being pulled up by a force of 4000N at an angle of: A) Up at 50 degrees. B) Down at an angle of 70 degrees.

2. Relevant equations
Fw = mg
Fsinx
Fcosx

3. The attempt at a solution

A) Fn = 784 - 4000sin50 = -2280 = 2280N

B) Fn = 784 + 4000sin70 = 4543N

Is this correct?

2. Feb 24, 2009

### septum

part B would have to be sin(-70) since it is below, correct?

3. Feb 24, 2009

### Intrusionv2

But did you notice that I added instead of subtracted the 2nd one?

4. Feb 24, 2009

### septum

Okay, well as I understand it for part A normal force is equal to mgcos50 degrees. I don't think you have to multiply anything but gravity by your mass.

Thus, n=mgcos: n=(80kg)(9.8 m/s^2)*cos(50)

5. Feb 24, 2009

### Intrusionv2

It is not on an incline. it is being pulled up at an angle.

6. Feb 24, 2009

### septum

I can't picture anything being pulled up at an angle without being on an incline so I'm probably not the person to be trying to help. :/ Sorry

7. Feb 24, 2009

### Sonolum

Have you tried drawing a free body diagram? What is the Normal force equal to? In part (a), is the pulling increasing or decreasing the normal force? In part (b) is the pulling increasing or decreasing the normal force?