Statics Problem: Weight Hanging from a Rope

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Homework Help Overview

The problem involves a mass hanging from a pulley system, specifically a 50 kg mass and a 1 kg mass, with the goal of understanding the relationship between the movements of these weights and the length of the rope. The discussion centers around the forces acting on the system and the geometry involved in the setup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equilibrium of forces in the system, particularly focusing on the equation "2*50sin(x)=1" and its implications. There are attempts to clarify the free body diagram (FBD) of the 1 kg weight and the forces acting on it. Some participants express confusion regarding the solution provided and seek clarification on specific steps and relationships in the problem.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and attempting to clarify their understanding of the forces involved. Some guidance has been provided regarding the relationship between the movements of the weights and the geometry of the rope, but there is no explicit consensus on the solution or final outcome.

Contextual Notes

Participants are working with a specific setup involving a pulley and two weights, and there is mention of a solution that some find difficult to follow. The problem is framed in terms of the change in height of the 50 kg mass and the corresponding change in the rope's length.

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Homework Statement


Mass hanging from the pulley in 50kg, the mass on the rope is 1kg. Please see attachment for details. The answer will be in terms on L.


Homework Equations


The sum of the forces in the y-direction and x-direction are equal to 0.


The Attempt at a Solution


Adding the moments at the pulley and equating them to 0; we have that T=50.
Now a free body diagram of the second weight. 2*50sin(x)=1, sinx=(1/100)=H/X
X=(L^2+H^2)^(1/2), substituding this in (1/100)=H/X, and solving for H, I find H=L/100.
Which is not the answer I have in the solutions.
Any help will be greatly appreciated.
 

Attachments

  • prob1.png
    prob1.png
    4.2 KB · Views: 576
  • staticproblem1.png
    staticproblem1.png
    8.1 KB · Views: 531
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Can you explain to me "2*50sin(x)=1"?
 
When I draw the FBD of the 1kg weight; I addthe forces in the y-direction
50*sin(x)+50*sin(x)=1
 
What is the answer that they give?
 
I have attached the solution. I cannot follow what they are doing; if anybody does please help.
 

Attachments

  • solutionprob5.png
    solutionprob5.png
    14.5 KB · Views: 554
The key here is that they are asking how much the 50kg weight will move up due to the 1kg weight. the Solution shows the change in length of the rope over the horizontal section, which is the same as the change in height of the vertical section, which is the same as the 50kg mass.
 
Ok, I understand that now. But I still don't get where the 50/L is coming from.
 
sin(tetha)=50/L
 
Ok got it, thanks for the explanation.
 

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