# Various questions, mechanics

1. Aug 9, 2007

### Niles

1. The problem statement, all variables and given/known data

A mail bag with m = 120 kg is suspended by a vertical rope 8 m long.

1) I have to find the force needed to hold the bag in it's position.

2) I have to find out, how much work is done by the worker in moving the bag to this position.

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3) When a trolley is being pulled up a ramp 4 m, and the ramp inclines 30 degrees, find the work done by the normalforce.

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4) When a rope tied to a body is pulled, the body accelerates. But the body excerts an equal, opposite, force - is the total work done 0?

3. The attempt at a solution

1) I've found F = m*g*tan(a), where a is the angle between the vertical line and the string suspending the bag.

2) I thought of integrating m*g*tan(a) with limits 0 and 30 (degrees, cause 30 degrees is when x = 4 m) and multiplying by 4 m, since W = F*s. Please confirm.

3) I believe it's m*g*cos(30) * 4 m * cos(120) (120 degrees, because of 90 + 0).

4) I believe it is. But it's only an assumption.

Last edited: Aug 9, 2007
2. Aug 9, 2007

### mgb_phys

1, You seem to have some extra info about an angle that we don't.
Can you include a picture?
2, We have to knwo where the bag started from.
3, Remember work = force * distance
4, If you were correct you would be able to pull heavy objects around all day with no effort!

3. Aug 9, 2007

### Niles

1) It's placed 4 metres from the center. Sin(a) = 1/2, so a = 30. Same figure as http://www.math.duke.edu/education/ccp/materials/diffeq/pendulum/pendulum.gif - but not same numbers.

2) The bag starts from the pole, and is being moved 4 metres.

3) I'm confused about the angle. A = F*s*cos(a). But is a 30 degrees, 90 or 120?

4) Oh yeah.. my bad :-)

But perhaps you can answer another question. When I pull a rope attached to a rock, I am being affected by an equal, opposite, force. Why isn't the tension in the rope zero?

Last edited: Aug 9, 2007
4. Aug 9, 2007

### mgb_phys

1, Depends on the direction of the force. The vertical component is F=mg the magnitude depends on the direction you are pulling in.
2, Just need absolute change in height * g - path doesn't matter
3, Is it the vertical componnet of the normal force * vertical distance + horizontal component * horizontal distance?

The opposing force is in the opposite direction and so adds to your force.

5. Aug 9, 2007

### Niles

1) He is pulling it from center to right. So the resultant_force is pointing to the center - and this is the force I've found.

2) So delta E_pot = W?

3) Hmm, let's do it another way. A box is placed on a ramp with an incline of 30 degrees (the angle is 30 with horizontal) and travels the distance 2 m on the ramp. How would you find the work of the normalforce?

6. Aug 9, 2007

### Niles

Last edited: Aug 9, 2007
7. Aug 9, 2007

### mgb_phys

Not sure what it means by the work done by the normal force!
Does it mean the work done against friction ( ie frcition*normal force * slope distance) or does it mean the vertical component of the normal force * height?
I've never seen this phrase before in a question.

8. Aug 9, 2007

### Niles

I don't know what it actually is - it's also the first time I've heard it. But what I can tell you is that it - apparently - equals zero everytime, because the angle is 90 degrees.

But regarding the second question:

Will the total work done, A_t, be delta E_pot? Or can I use my integration-trick?

Last edited: Aug 9, 2007