Two Normal Forces: Physics Explanation

[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, a lot of things i have learned in physics are contrary to common knowledge.

 

[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.

 

[charng]

Hello,

Great question! In this scenario, the box is experiencing two normal forces because it is in contact with two surfaces - the left and right sides of the ramp. In this case, the friction force would indeed be equal to the sum of the two normal forces multiplied by the coefficient of friction (Ff = (FnL + FnR) * mu). This is because the friction force is a result of the two surfaces rubbing against each other, and the magnitude of this force depends on the normal force and the coefficient of friction.

However, it's important to note that the normal forces in this scenario would not necessarily be equal to half of the weight of the box. This is because the weight of the box is acting downwards, while the normal forces are acting perpendicular to the surfaces. The magnitude of the normal force would depend on the angle of the ramp and the weight of the box.

You are correct in thinking that the box being encased in the ramp would make it slide less easily compared to a box on only one side. This is because the friction force is directly proportional to the normal force, so having two normal forces would increase the friction force and make it more difficult for the box to slide.

I hope this explanation helps. Keep asking questions and exploring the world of physics!