# Homework Help: Calculate the magnitude in newtons of the normal force

1. Nov 5, 2007

### anap40

1. The problem statement, all variables and given/known data
A plank, of length L = 3.7 m and mass M = 7.0 kg, rests on the ground and on a frictionless roller at the top of a wall of height h = 1.70 m (see the figure). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ >= 74° but slips if θ < 74°. Calculate the magnitude in newtons of the force exerted by the roller on the plank when θ = 74°.

I got this part, the force is 19.8N

Continuation: Calculate the magnitude in newtons of the normal force exerted by ground on the plank when θ = 74°.

I get stuck here.

2. Relevant equations
T=Ia

3. The attempt at a solution
I calculated the torque down that goes thrugh the center of mass of the plank.( the center of mass is just above the corner of the ledge that the plank is leaning on)
I also calculated the torque that the edge of the wall exerts on the board in the opposite direction.

If I set the point where it contacts the ground as the pivot point, I get
T(of COM) - t(of wall) = 0

The only problem that the above is not true. I know that I am missing something but I am not sure what it is.

Last edited: Nov 5, 2007
2. Nov 5, 2007

### anap40

I forgot to add the picture, its preatty much impossible to do w/o the pic.

http://img237.imageshack.us/img237/1950/prob09aur2.gif [Broken]

Last edited by a moderator: May 3, 2017
3. Nov 5, 2007

### Staff: Mentor

Don't forget the conditions for translational equilibrium. In particular, the sum of the vertical forces must be zero.
Why do you think it's not true? Since the plank is in equilibrium, torques about any point must equal zero. (It might not be helpful, but it's true.)

4. Nov 5, 2007

### anap40

But the only vertical forces are the forcce of gravity and the normal force right?

5. Nov 5, 2007

### anap40

I also went back and checked and T(COM) - T (of wall) does equal 0.

But I am still stuck, I cant simply say that the normal force is equal to the weight because, the torque from the COM is will cause the normal force to be less that that, but I am not sure how to figure out how much less

6. Nov 5, 2007

### Staff: Mentor

No. The force exerted by the roller also has a vertical component.

Good.

7. Nov 5, 2007

### anap40

Thanks again doc, for some reason I just figured that the roller was only a horizontal component. It turns out that the force is actually perpendicular to plank.