Calculating work when 2 forces are present

In summary, the conversation is discussing a problem involving a man pushing a 75 kg crate at a constant velocity of 225 N at a downward angle of 15° for a distance of 12 m. The coefficient of friction between the crate and the floor is 0.24. The question is how much work did the man do on the crate. The conversation goes through various attempts at solving the problem, with one person suggesting that the work should only take into account the force applied by the man, while another suggests taking into account both friction and the man's force. The final answer given is 320N-m.
  • #1
jono240
11
0

Homework Statement


This should be an easy one. Please tell me what I am 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

Homework Equations



Ffriction = μ*Fn
W = F*Δd*cosθ

The Attempt at a Solution


First attempt
Ffriction = 176.4N
So
W = F*Δd*cosθ
W = (225 - 176.4)*12*cos15
W = 563.33N/m (wrong answer)

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
W = 491.2N/m (wrong answer)

I've tried so many combinations I can't seem to figure it out. The answer is 320N/m btw
Please tell me what I am doing wrong
 
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  • #2
The thing is, it's the work the man is doing on the crate. So why is friction relevant?

It should just be

[itex]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[/itex]

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
The answer is 320N/m btw

Would you mean N-m for work.
 
  • #4
ya n-m

does anyone have an answer for this though?
I found a question from the textbook that had a wrong answer in the back so I don't care about that answer anymore. I am starting to think my first attempt is the correct way. Can anyone confirm that for me please?
 
  • #5
^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
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
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!
 

1. How do you calculate work when there are 2 forces present?

To calculate work when there are 2 forces present, you will need to use the formula W = F1 * d1 * cosθ1 + F2 * d2 * cosθ2, where F1 and F2 are the magnitude of the forces, d1 and d2 are the displacement in the direction of the force, and θ1 and θ2 are the angles between the force and the displacement.

2. What if the 2 forces are acting in opposite directions?

If the 2 forces are acting in opposite directions, you will need to subtract the smaller force from the larger force in the formula. For example, if one force is 10 N and the other is 5 N, you would use (10 N - 5 N) for the force in the formula.

3. Can you calculate work if the forces are not acting in the same direction?

Yes, you can still calculate work even if the forces are not acting in the same direction. However, you will need to use the component of the force that is in the same direction as the displacement in the formula. This is represented by the cosθ value.

4. How do you determine the angle between the force and the displacement?

The angle between the force and the displacement can be determined by using the trigonometric function cosine (cos). You will need to use the angle between the force and the displacement in the formula for calculating work.

5. What are the units for work when there are 2 forces present?

The units for work when there are 2 forces present are the same as for regular work, which is joules (J). This is because work is a measure of energy, and energy is measured in joules.

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