Force required to pull trunk across floor

[Ace.]

Homework Statement

You are dragging a 110 kg trunk across a floor at a constant velocity with horizontal force of 380 N. A friend decides to help by pulling on the trunk with a force of 150 N [up]. Will this help? Calculate the force required to pull the trunk at a constant velocity to help you decide.

Homework Equations

μ_k = F_k/F_N
F_g = m × g

The Attempt at a Solution

I found the coefficient of kinetic friction (μ_k) to be 0.35.

I'm confused by what it means by "will this help?". Wouldn't it help because your friend is reducing the F_N (force normal) on the trunk? I found force required to pull the trunk at a constant velocity is 377.3 N (F_k) using the equation:

μ_k = F_k/F_N
= 0.35 x 1078 N
= 377.3 N.

Now, when your friend is pulling the trunk upwards wouldn't the Force normal decrease to 928 N [up] because 1078 N - 150 N? So would the force required be

F_k/ F_N = 324.8N?

 

[Simon Bridge]

I found the coefficient of kinetic friction (μ_k) to be 0.35.

What is important is how you found it. My number may agree with yours but if I got it by furvent prayor I probably won't get the marks.

I'm confused by what it means by "will this help?". Wouldn't it help because your friend is reducing the F_N (force normal) on the trunk?

Well would it? If so, then it does help and if not then it doesn't. Where is the confusion?

Note: someone not familiar with the way friction works may think that people who help should pull, at least a bit, in the same direction as you.

Anyhow - reading the rest - your reasoning seems fine if a little disorganized.
One way to overcome confusion and uncertainty with this sort of problem is to draw the free body diagram (or some other reasonable diagram) with all the forces and formally write ƩF=ma for each axis direction.

So : constant velocity implies acceleration is zero so ma=0 ... and you can write:
- by yourself: $\sum F=ma \Rightarrow$
vertically: $F_N-mg=0$
horizontally: $F_{me}-\mu F_N=0$

- with your friend: $\sum F=ma \Rightarrow$
vertically: $F_N+F_{him}-mg=0$
horizontally: $F_{me}-\mu F_N=0$

See how it is easier to have confidence in your results when it is written like that?