How Do You Calculate the Other Student's Force on a Trunk in Physics?

[Balsam]

Homework Statement

Two students push horizontally on a large, 65kg trunk. The trunk moves east with an acceleration of 2.0m/s^2. One student pushes with a force of 2.2x10^2 N[E 42 degrees S]. The force of friction acting on the trunk is 1.9x10^2N[W]. Determine the force that the other student applies on the trunk.Looking at the solutions manual, you solve this by finding Fnet (you do Fnet=ma). Then you do (Fnet)x = Ff+(Fa1)x+(Fa2)x. You plug in everything but (Fa2)x-- solve for (Fa2)x. For the y components, you do (Fnet)y=(Fa1)y+(Fa2)y, and you solve for (Fa2)y. However, the y component equation does not include Fn or Fg. Why is this? You could say that the normal and gravitational forces cancel out, but since we know one of the applied forces is 2dimensional, it has a y component too, meaning Fn and Fg don't cancel out since they're not the only 2 vectors with y components. Can someone please explain why Fn and Fg weren't included in these calculations.

Homework Equations

I already stated the solution above.

The Attempt at a Solution

Is the reason that Fn and Fg were left out of the calculations because you cannot solve for Fn with the given information?[/B]

 

[SteamKing]

Homework Statement

Two students push horizontally on a large, 65kg trunk. The trunk moves east with an acceleration of 2.0m/s^2. One student pushes with a force of 2.2x10^2 N[E 42 degrees S]. The force of friction acting on the trunk is 1.9x10^2N[W]. Determine the force that the other student applies on the trunk.Looking at the solutions manual, you solve this by finding Fnet (you do Fnet=ma). Then you do (Fnet)x = Ff+(Fa1)x+(Fa2)x. You plug in everything but (Fa2)x-- solve for (Fa2)x. For the y components, you do (Fnet)y=(Fa1)y+(Fa2)y, and you solve for (Fa2)y. However, the y component equation does not include Fn or Fg. Why is this? You could say that the normal and gravitational forces cancel out, but since we know one of the applied forces is 2dimensional, it has a y component too, meaning Fn and Fg don't cancel out since they're not the only 2 vectors with y components. Can someone please explain why Fn and Fg weren't included in these calculations.

Homework Equations

I already stated the solution above.

The Attempt at a Solution

Is the reason that Fn and Fg were left out of the calculations because you cannot solve for Fn with the given information?[/B]

You are given the force of friction (190 N) which opposes the motion of the trunk.

Since the force of friction is already given, what would be the point of including the normal or gravitational forces acting on the trunk?

Remember, these two forces, Fn and Fg, act perpendicular to the motion of the trunk, so their presence is not going to affect the motion of the trunk across the floor, insofar as friction is concerned, the force of which is already given.

 

[Balsam]

You are given the force of friction (190 N) which opposes the motion of the trunk.

Since the force of friction is already given, what would be the point of including the normal or gravitational forces acting on the trunk?

Remember, these two forces, Fn and Fg, act perpendicular to the motion of the trunk, so their presence is not going to affect the motion of the trunk across the floor, insofar as friction is concerned, the force of which is already given.

But the applied force has an x and y component. Shouldn't you include all the y component vectors in calculating the y component of the applied force?

 

[SteamKing]

But the applied force has an x and y component. Shouldn't you include all the y component vectors in calculating the y component of the applied force?

But the pushing forces and the friction force are all occurring in the x-y plane. The normal force and the gravity force are still acting normal to the x-y plane, parallel to the z-axis if you will.