# Forces and acceleration of pulled rope

1. Sep 22, 2006

### trajan22

A man with mass 70.0 kg stands on a platform with mass 25.0 kg. He pulls on the free end of a rope that runs over a pulley on the ceiling and has its other end fastened to the platform. The mass of the rope and the mass of the pulley can be neglected, and the pulley is frictionless. The rope is vertical on either side of the pulley.

With what force does he have to pull so that he and the platform have an upward acceleration of 1.80m/s^2

ok so my force diagram shows 2 forces acting on this firstly there is the force of mg pointing down. then there is the force of acceleration in the upward direction and this is equal to a*g.

so the way i figured the problem is that mg+ag=T wher T is the tension in the rope. the reason that i got this equation is because the force needed to pull the platform and person up must be the acceleration upward plus the gravity times the mass. However i dont think this reasoning is right so any help on this would be helpful. thanks

2. Sep 22, 2006

### Staff: Mentor

The weight of "man+platform" does pull down. What forces pull up?
Force of acceleration? Not sure what that means!

What you need to do is to draw a diagram showing all forces acting on "man+platform". Then apply Newton's 2nd law: set the net force equal to m*a.

Hint: Which end of the rope pulls up on the "man+platform" system?

3. Sep 22, 2006

### trajan22

sorry i dont have a scanner so ill try to explain the force diagram i drew.
The F=mg points down
and the tension in the rope points up
i have no normal force because the platform is held up by the tension in the rope...
and in answer to your hint question the end that is attached to the platform pulls up.

are these ideas right?

4. Sep 22, 2006

### Staff: Mentor

OK. I will assume you are taking as the "man + platform" as your system and thus are analyzing all the forces that act on that system.
OK.
OK.
There is a normal force between platform and man (and between man and platform), but since you are considering them as a single system they are internal forces and don't affect the net force.

What about the other end where the man is pulling? If the man is pulling on the rope, what does Newton's 3rd law tell you?

5. Sep 22, 2006

### trajan22

newtons third law says that the rope is also pulling on the man in equal and opposite direction...
however im still unsure as to where to go from here this is all ive been able to come up with but i know its wrong
T=m(g+a) then m(g+a)=mg?

Last edited: Sep 22, 2006
6. Sep 22, 2006

### trajan22

i just wanted to keep this thread updated that im still confused on this problem??

7. Sep 23, 2006

### trajan22

ok so ive still been working on this same problem but im still only able to find it as T=m(g+a) because these are the only forces acting on this object something is wrong though because this isnt the right answer.

8. Sep 23, 2006

### trajan22

actually i just figured it out...haha thanks anyway

9. Sep 24, 2006

### Staff: Mentor

The thing to realize is that the upward force is 2T, not just T, since both ends of the rope pull up. The downward force is the weight, mg. Set the net force equal to ma (per Newton's 2nd law) and solve for T--since the man must exert a force on the rope equal to its tension.