For part A, the question asks to determine the normal component of force of the path acting on particle p. Is this the normal force or the sum of forces acting in the e_n direction?
We are looking for the component of the contact force of the surface on the particle acting in the direction perpendicular to the path. It should include no other forces.
Just to ensure I am conceptually sound, in this problem the path is a U shape. Thus the en unit vector will point inward towards the center of the U, in the same direction as the normal force vector, correct?
Is it possible to have a negative normal force? Would you use the universal vectors or make e_t and e_n the positive direction? When I did it the latter way, I found a negative normal force, which doesn't seem to make sense?
A negative sign on any reaction force for which you solve in the problem means one thing: your calculations indicate that the force is in the opposite direction from what you assumed when you drew the FBD. There are several reasons that the sign could end up negative:
* You simply made a mistake when you drew the FBD.
* You made algebra errors somewhere.
* You did not have enough insight to figure out the direction of the force at the start of the problem, so you guessed. The negative sign said that you guessed incorrectly. Generally no big deal.
* We will see a number of problems this semester of particles sliding over curved surfaces. If the speed of the particle is too fast, the particle could lose contact with the surface. So, if the normal force ends up negative, it would mean that the particle lost contact (that is, the surface would have to "pull back" on the particle to keep things in contact).
I'm a little confused on the g arrow shown to the right of the picture. The way I understand a horizontal plane is that gravity acts perpendicular to the plane, and would therefore not matter, however the g arrow looks like gravity is shown in the negative y, should we include gravity in our calculations?
You are correct - the downward arrow for gravity is contradictory to the label of HORIZONTAL PLANE.
* If you have not worked the problem yet, work it as being in the VERTICAL PLANE, since the solution for the problem is based on that.
* If you have already worked the problem and used HORIZONTAL PLANE, then no need to re-work. We will count either solution as being correct.
I solved for the Normal force in two different ways, first I used, the sum of the Ft forces, since N*u_k is the force of friction againts mgcostheta. However, when I solved for the normal force for a second time with the Fn equation, just mgsintheta - N, I got a completely different answer. I was just wondering which one of these methods was correct, or if they should yield the same result.
For part A, the question asks to determine the normal component of force of the path acting on particle p. Is this the normal force or the sum of forces acting in the e_n direction?
We are looking for the component of the contact force of the surface on the particle acting in the direction perpendicular to the path. It should include no other forces.
Just to ensure I am conceptually sound, in this problem the path is a U shape. Thus the en unit vector will point inward towards the center of the U, in the same direction as the normal force vector, correct?
Is it possible to have a negative normal force? Would you use the universal vectors or make e_t and e_n the positive direction? When I did it the latter way, I found a negative normal force, which doesn't seem to make sense?
A negative sign on any reaction force for which you solve in the problem means one thing: your calculations indicate that the force is in the opposite direction from what you assumed when you drew the FBD. There are several reasons that the sign could end up negative:
* You simply made a mistake when you drew the FBD.
* You made algebra errors somewhere.
* You did not have enough insight to figure out the direction of the force at the start of the problem, so you guessed. The negative sign said that you guessed incorrectly. Generally no big deal.
* We will see a number of problems this semester of particles sliding over curved surfaces. If the speed of the particle is too fast, the particle could lose contact with the surface. So, if the normal force ends up negative, it would mean that the particle lost contact (that is, the surface would have to "pull back" on the particle to keep things in contact).
Does this help?
I'm a little confused on the g arrow shown to the right of the picture. The way I understand a horizontal plane is that gravity acts perpendicular to the plane, and would therefore not matter, however the g arrow looks like gravity is shown in the negative y, should we include gravity in our calculations?
You are correct - the downward arrow for gravity is contradictory to the label of HORIZONTAL PLANE.
* If you have not worked the problem yet, work it as being in the VERTICAL PLANE, since the solution for the problem is based on that.
* If you have already worked the problem and used HORIZONTAL PLANE, then no need to re-work. We will count either solution as being correct.
Apologies for this error.
I solved for the Normal force in two different ways, first I used, the sum of the Ft forces, since N*u_k is the force of friction againts mgcostheta. However, when I solved for the normal force for a second time with the Fn equation, just mgsintheta - N, I got a completely different answer. I was just wondering which one of these methods was correct, or if they should yield the same result.
Can this be solved in a cartesian coordinate system rather than path?