What Happens When a Monkey Pulls on a Rope with a Weight on the Other Side?

  • Thread starter Thread starter borku1
  • Start date Start date
  • Tags Tags
    Mechanics
AI Thread Summary
When a monkey pulls on a rope with a weight on the other side, the tension in the rope increases, affecting both the monkey and the weight. The tension is equal on both sides of the pulley, meaning both the monkey and the 5 kg weight experience the same force. As the monkey pulls down, it can either rise off the floor, the weight can rise, or both can rise at the same rate depending on the forces involved. The discussion emphasizes using Newton's second law (F = ma) to analyze the forces acting on both the monkey and the weight. Understanding the tension and gravitational forces is crucial to determining the outcome of this scenario.
borku1
Messages
2
Reaction score
0

Homework Statement


A weightless rope is draped over a frictionless pully attached to the ceiling (change of direction only, no mechanical advantage). Attached to one end of the rope is a 5 kg weight sitting in the floor. A 5 kg monkey grabs the other end of the rope and begins to pull down on the rope.
Does:
a) the monkey rise off the floor,
b) the 5 kg weight rise up off the floor, or
c) both the monkey and the weight rise off the floor at the same rate.


Homework Equations


This is a question from a mechanics section of the high school physics book i had in the early 60's (algebra & trig, no calc). Since this question has stuck with me all these years, i would appreciate a pointer to what concept or equations i should consider.


The Attempt at a Solution



thanks in advance
 
Physics news on Phys.org
Welcome to PF!

Hi borku1! Welcome to PF! :smile:

Hint: if the monkey is pulling down, the tension in the rope will increase. :wink:
 
i agree, and i believe the tension would be the same on the 5kg monkey and 5kg weight. For instance, after the monkey first establishes 5kg tension on the rope, and then, by further pulling on the rope, shortens it by 2 meters, one or both of the monkey and weight will be a total of 2 meters up off of the floor. And i believe the potential energy of either end system (monkey up 2 meters, weight up 2 meters, or both up 1 meter) will be the same.

The force the monkey applies to the rope could equally:
accelerate the monkey's mass by 'x', or
accelerate the weight's mass by 'x', or
accelerate both his mass and the weight's mass (10kg) by 1/2 'x' (half the acceleration), moving both off the floor an equal amount.

so, i guess i miss the point of tension as a hint. How would the tension lead us to select one scenario over the other?

thanks again in advance
 
The tension of the rope and the force of gravity are the only forces on the monkey, and on the weight, since the tension is the same on both sides of the pulley ...
 
borku1 said:
… How would the tension lead us to select one scenario over the other?

Call the tension T (as willem2 :smile: says, it's the same on both sides of the pulley), and do F = ma separately for the monkey and for the 5 kg weight. :wink:

What do you get?​
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top