Popeye pulling himself up using a pulley

[totallyclone]

Popeye was lifting himself to the top of a 30m mast until he was 10 m away from the pulley. The rope was 100m long and weighed 50kg and he would pull himself off of the ground using a slow constant speed. How hard was he pulling when he was 10m below the pulley?

 

[MostlyHarmless]

Good question. Tell us what you think!

As per the rules of the forum, you need to show some attempt at solving this.

 

[totallyclone]

I think that the forces acting on the system would be (when you draw a free body diagram of the rope) that when looking at the side at which Popeye is on the chair there is a force of gravity going downward and a force of tension going upward on and the force being applied by Popeye to hold himself up at 10m. However on the right side where Popeyes's chair is not there is a force of gravity going downward and a force of tension going upward. Then I would find the vector summation in the y axis.

 

[MostlyHarmless]

First, you should understand that something cannot "weigh" 50"kg". Weight is a force due to gravity, whereas kg is a unit of mass. So the rope has a mass of 50kg and has a weight of mg.

With your free body diagram,
1.) The side that popeye is sitting on you'll have tension pulling up and the Weight(Force due to gravity) of Popeye pulling down and the weight of 10m of roping pulling down.

2.)The other side will have Tension pulling up and the force at which popeye is pulling down and the weight of 30m of rope pulling down.

*30m of rope because, even though there is only 10m of rope on the other side, since the mast is only 30m, 60m of rope will be on the ground, and 30m will be hanging.*

Now apply Newtons 2nd law, and set up your equations for each free body diagram.
He is ascending at a constant speed, so we know that a=0.

See how far you can get from here let us know.

 

[totallyclone]

Alright so this is the diagram I came up with:

And this is the solution I've worked out:

 

[MostlyHarmless]

I noticed you never said what the mass of popeye was, so I can't say for sure if you're right, so I'll just show you how I would set it up.
I'll call the Mass of of the rope supporting popeye, R1 and the other side of the rope R2. T is the tension, and M1 is popeye. and the Force applied by popeye $$F_{app}$$

So from our first free body diagram, we get $$F_{net}=0=T-(m_1g+r_1g)$$ and from the second we get we get $$F_{net}=0=(r_2g+F_{app})-T$$ Now we have a system of equations that we can solve by just adding them together. $$r_2g+F_{app}-m_1g-r_1g=0$$ Solving for $$F_{app}$$ we get $$F_{app}=m_1g+r_1g-r_2g$$
This should yield the answer you are looking for.

We know that popeye is moving up, so in order for him to move up T has to be greater than the forces pulling down on popeye. So that is why we do $$T-(m_1g+r_1g)$$
And similiarly the rope is moving down on the other side, so we know the weight of the rope plus the force popeye is pulling down at has to be greater than tension. So $$(r_2g+F_{app})-T$$

 

[MostlyHarmless]

I attached a picture of my diagram and equations.

The right diagram is popeye moving up, the right is the rope being pulled down.

 

[totallyclone]

I noticed you never said what the mass of popeye was, so I can't say for sure if you're right, so I'll just show you how I would set it up.
I'll call the Mass of of the rope supporting popeye, R1 and the other side of the rope R2. T is the tension, and M1 is popeye. and the Force applied by popeye $$F_{app}$$

So from our first free body diagram, we get $$F_{net}=0=T-(m_1g+r_1g)$$ and from the second we get we get $$F_{net}=0=(r_2g+F_{app})-T$$ Now we have a system of equations that we can solve by just adding them together. $$r_2g+F_{app}-m_1g-r_1g=0$$ Solving for $$F_{app}$$ we get $$F_{app}=m_1g+r_1g-r_2g$$
This should yield the answer you are looking for.

We know that popeye is moving up, so in order for him to move up T has to be greater than the forces pulling down on popeye. So that is why we do $$T-(m_1g+r_1g)$$
And similiarly the rope is moving down on the other side, so we know the weight of the rope plus the force popeye is pulling down at has to be greater than tension. So $$(r_2g+F_{app})-T$$

Oh I forgot to say what Popeye and the chair's mass was. Together, Popeye and the chair is 50kg.

 

[totallyclone]

Note for this question too that I missed: Force of tension is not the same.

Let me try this now see what I come up with.

 

[MostlyHarmless]

Ok, so then if the rope has a mass of .5 kg per 1 meter. The way you did it you have that the rope is 2m/kg. But what you want is kg/m so that wehn you multiply by m you are left with kg. so 50kg/100m=.5kg/1m

 

[MostlyHarmless]

Yeah, the tension on both sides of the rope are going to be the same. I'm having a hard time following your diagram. Seems like you're over thingking it a bit.

 

[totallyclone]

Yeah, the tension on both sides of the rope are going to be the same. I'm having a hard time following your diagram. Seems like you're over thingking it a bit.

Alright so this is what I came up with and found the F_A by Popeye is 294N? I sort of changed the diagram by a slight bit because the first diagram that you seen was my first attempt on this question.

 

[MostlyHarmless]

You just made an arithmatic mistake, otherwise it looks good! You did, 490-49, but its -490-49. Final answer should be 392N

 

[totallyclone]

You just made an arithmatic mistake, otherwise it looks good! You did, 490-49, but its -490-49. Final answer should be 392N

Wow I didn't see that. Well, now that I fixed it... thank you so much for pointing me in the right direction and spotting my mistakes! Cheers! :)

 

[MostlyHarmless]

You're more than welcome! Fun problem.