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Determine the acceleration of rope hoist

  • Thread starter gameguru
  • Start date
10
0
Problem:

To hoist himself into a tree a 72 Kg man ties one end of a nylon rope of negligible weight around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of 358N. Neglect any friction between the rope and the branch and determine the accleration.

Solution

The only force we need concern ourselves with seems to be the vertical motion. A free body diagram shows a downward force of mg, the rope the man pulls on (T1) and an upward force that the rope exerts on the man based on Newtons 3rd law (T2). My dilema is that if the man pulls on the rope, T1, then doesn't part of his weight (mg)get distributed to the force of T1 since they are both in the same direction. Once the man is off the ground then the full downward force would = T1 + mg. Since no friction occurs between the rope and the branch, the sum of the forces T1,mg would = T2.
Sum of all forces = T2 = ma and solving for a= T2/m. Is this the right approach or am I missing something?.
 
Last edited:

Answers and Replies

1,036
1
I can't exactly follow your reasoning, but here's how I'd approach it.

The only source of tension in the rope is the force that the man applies to it. Think, how much tension would there be if he let go? So T = 358 N. But there are two "legs" to the rope, one pulling up on his waist and one pulling his hands. So the total upward force on the man is 2T. The only downward force is mg. 2T + mg = ma.
 
HallsofIvy
Science Advisor
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Originally posted by Gameguru My dilema is that if the man pulls on the rope, T1, then doesn't part of his weight (mg)get distributed to the force of T1 since they are both in the same direction.
No, he is pulling down on one end of the rope with force T1. His weight is pulling down on the OTHER end of the rope. because of the "pully" (the turn around the limb) the forces on the man are in opposite directions. While the man is pulling down on one end of the rope, the rope goes around the tree limb so that force is upward on the man. The net force on the man is T1- mg.

Since the man's mass is 72 kg, his weight is 72*9.8= 705.6 Newtons. He is pulling down with force T1= 358 Newtons? He isn't going to go anywhere!
 
1,036
1
The net force on the man is T1- mg.

Since the man's mass is 72 kg, his weight is 72*9.8= 705.6 Newtons. He is pulling down with force T1= 358 Newtons? He isn't going to go anywhere!
I don't think so.

If that were true, a 180 lb. man on a 70 lb. scaffold would have to exert more than 250 lb. force to raise himself and the scaffold using a rope and pulley. I don't think most painters can do that.

In gameguru's problem, T pulls upward on the man's hands and on his waist. The only downward force on the man is his weight. So (taking UP to be positive) T = 358 N, mg = -705.6 N.

2T + mg = 716 - 705.6 = 10.4 = ma
a = 10.4/72 = 0.144 m/s^2 upward
 
508
0
How about using energy conservation to solve this? I like energy conservation.

Work invested into the system by man:
W = F * 2h (factor 2 because of pulley)
Where F is force, h is height.
Potential energy of system:
T = mgh
Where m is mass of man.
Kinetic energy of system:
V = mv2/2
Where v is velocity.

Conservation of energy:
W = T + V
2Fh = mgh + mv2/2
Solve!
v2 = (4F/m - 2g)h
Differentiate!
2va = (4F/m -2g)v
v cancels!
a = 2F/m -g.

I agree with gnome.
 
10
0
Thanks everyone for the feedback.

The answer appears to be .14 m/s2. Looks like my initial confusion was coming from the fact that the man's weight wa attached to both ends of the rope!

Thanks again.
 

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