# Magnitude of Normal Force (Frictionless)

1. Sep 18, 2008

### Spartan Erik

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

"A crate of mass 50 kg is pushed across a frictionless horizontal floor with a force of 100N directed 23.5 degrees below the horizontal. The magnitude of the normal force of the floor on the crate is:"

2. Relevant equations

F = ma? Not sure what else would apply.. mainly a conceptual issue here

3. The attempt at a solution

I took 100N x cos(23.5) in order to get the horizontal force which turns out to by 91.7N.. not sure what to do from here

2. Sep 18, 2008

### Staff: Mentor

To solve for the normal force, analyze the vertical components of the forces acting on the crate. (Three forces act.)

3. Sep 18, 2008

### Spartan Erik

Well gravity is acting on the crate (9.8 m/s^2 downward), and the crate itself has a normal force exerted in the opposite direction

4. Sep 18, 2008

### Staff: Mentor

Gravity and the normal force are two of the three forces. What's the third vertical component?

5. Sep 18, 2008

### Spartan Erik

Well I imagine the third component could be sin(23.5 degrees) x 100 = 39.875

6. Sep 18, 2008

### Staff: Mentor

Right. So what must the normal force be to balance this force plus gravity?

7. Sep 18, 2008

### Spartan Erik

Ah so 9.8 m/s^2 x 50kg = 490N

And adding that to 39.875N will result in a magnitude of 530N