# Two normal forces?

1. Nov 5, 2008

### lax1113

Hey guys,
So we haven't gotten to 3-d objects yet, but i was wondering, since Ff=Fn*mu, what would happen if an object was in a sort of like v, so that it had two normal forces.

Lets say it is a box, sliding down a 30degree ramp. The ramp is not just a ramp, but is folded so that it makes a 90 degree angle where the box rests. Because it is 90 degrees, the box is against the surface not only once, but twice, once on the left side of the box, once on the right. So, now that there are two normal forces, which have to be half of the mg i would guess, since the box is not accelerating in the y direction, would the friction also be half? Or is friction eqaul to the sum of Fnormal forces multiplied by Mu.

When i think about it just without applying any physics knowledge, it seems like a box being almost encased in a ramp, well at least on two sides, would slide less easily as a box only on one side. Then again, alot of things i have learned in physics are contrary to common knowledge.

2. Nov 6, 2008

### ZikZak

Yes, you would have two normal forces. No, they will not be each equal to 1/2 mg.

A "normal force" is simply a force of contact normal (or perpendicular) to the plane of contact. So if there are two points of contact, you will have two normal forces. In your example, the first normal force is oriented 30 degrees from vertical, becausae the box is on a 30 degree ramp. The second will be oriented 60 degrees from vertical, because the second ramp against which the box is jammed is a 60 degree ramp. If friction is involved, then there are two separate coefficients of friction, one for each ramp, and the forces of friction are Mu*Fnormal, one for each normal force, and each directed parallel to its ramp.

All four forces: two normal forces, and two frictional forces, have vertical components. You would have to set up $\Sigma F_y = 0$ and $\Sigma F_x=0$ and solve the equations to find what each normal force is.