# Calculating work when 2 forces are present

1. Oct 3, 2011

### jono240

1. The problem statement, all variables and given/known data
This should be an easy one. Please tell me what im missing.

A man is pushing a 75 kg crate at constant
velocity a distance of 12 m across a warehouse.
He is pushing with a force of 225 N at an angle
of 15° down from the horizontal. The coefficient
of friction between the crate and the floor is
0.24. How much work did the man do on the
crate?

Given - m=75kg, Δd=12m, Fa=225N [15° down from hor], μ=.24

2. Relevant equations

Ffriction = μ*Fn
W = F*Δd*cosθ

3. The attempt at a solution
First attempt
Ffriction = 176.4N
So
W = F*Δd*cosθ
W = (225 - 176.4)*12*cos15

Second attempt
W = F*Δd*cosθ
W = 225*12*cos15
W = 2608N/m

W = F*Δd*cosθ
W = 176.4*12*cos180
W = -2116.8N/m

W = 2608 - 2116.8

I've tried so many combinations I cant seem to figure it out. The answer is 320N/m btw
Please tell me what im doing wrong

2. Oct 3, 2011

### AuraCrystal

The thing is, it's the work the man is doing on the crate. So why is friction relevant?

It should just be

$W_{man}=\int \mathbf{F}^{man} \cdot d\mathbf{r} = \int (F^{man}_{x},F^{man}_{y}) \cdot (dx, dy) =\int F^{man}_{x} dx=F^{man}_{x} \Delta x= F^{man} \cos \theta \Delta x$

I computed that and got 2608 N/m. Maybe they meant the work done by both friction and the man. I also computed that but got 2525.08 N/m. Anyone else have any ideas? Am I just being an idiot? xD

Perhaps the answer in the back of the back is wrong. Maybe you should check with your prof.

3. Oct 3, 2011

### 256bits

Would you mean N-m for work.

4. Oct 3, 2011

### jono240

ya n-m

does anyone have an answer for this though?
I found a question from the text book that had a wrong answer in the back so I dont care about that answer anymore. Im starting to think my first attempt is the correct way. Can anyone confirm that for me please?

5. Oct 4, 2011

### AuraCrystal

^Thing is, it's the work the MAN does on the crate. You just take the force from him into account, nothing else. So I think the second one is correct.

6. Oct 4, 2011

### SammyS

Staff Emeritus
Total work done by the man minus the work he does in overcoming friction. That is 321.3 N-m . (Using g =9.91 m/s2)

Don't forget, when computing normal force that there is a component of the 225 force that's downward in addition to the weight of the box, mg. Therefore, the normal force must be equal but opposite to the sum of those downward forces.

7. Oct 4, 2011

### Staff: Mentor

Hmm. If the man is pushing the crate at a constant velocity, how is it that the horizontal component of the force he applies is greater than the friction force? Something got by the editors!