3 weights suspended on 2 pulleys to form a M shape

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Homework Help Overview

The problem involves a system of three weights suspended on two pulleys, forming an "M" shape. A 25 kg mass is positioned between two pulleys, each connected to 15 kg masses. The goal is to determine how far below the pulleys the 25 kg mass must be for the system to achieve equilibrium, given that the pulleys are 45 cm apart.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating tensions in the ropes and the forces acting on the system. There are attempts to apply Newton's laws and geometry to analyze the forces and angles involved. Some participants express uncertainty about identifying forces correctly and finding the necessary angles for calculations.

Discussion Status

The discussion is active, with participants exploring different aspects of the problem, including the application of Newton's first law and the geometric relationships in the system. Guidance has been offered regarding the use of vertical components of tension and the application of trigonometry, but no consensus has been reached on the specific approach to take.

Contextual Notes

Participants note the challenge of determining angles and components due to limited information about the geometry of the setup. There is an acknowledgment of the need to correctly identify forces and their components to proceed with the solution.

lksmith17
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Homework Statement


A block with a mass of 25 kg is hung midway between two pulleys, with a rope connecting a hook at the top of the block to each of the two pulleys. On the other side of the pulleys, both ropes are connected to blocks with masses of 15 kg. The 25 kg mass in the middle sags downward until the 3 blocks and 2 pulleys form the shape of an “M”. If the pulleys are 45 cm apart, how far below the level of the pulleys does the top of the 25 kg mass need to be in order for the system to be in equilibrium?


2. Homework Equations

Sum of Forces =0

3. The Attempt at a Solution
I have calculated the tensions in one rope to be 147N using 15kg*9.8m/sec^2 and the Fg of the middle block is 245N I know I need to find the angles of the rope to the ceiling by splitting the tension of the rope in x and y directions, but I can not figure out how to get those values.
 
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You know the tension in the rope, and you know the weight of the block in the middle. At the joint where they all meet, identify the forces acting in the y direction and use Newton 1. Then it's geometry.
 
At the joint where they all meet is at the middle block right? If so, the Force in the y direction 245N down from gravity but the tension is is from the two ropes is the force that counters the force of gravity. So I can make a right triangle, but i only have the hypotenuses' length. I don't have any other angles to figure out the the angle between the ceiling and the roof. I guess I am missing something completely could you let me know what I am doing wrong. Am I not identifying a force correctly?
 
From Newtons 1st law in the y direction, the sum of the vertical components of the tension forces in the ropes must equal the weight of the center block. So what is the sum of those tension components, and what is the vert component of the tension in one of the ropes. The tension force acts along the diagonal, so use Pythagoras to get the horiz comp. then do some geometry and trig to find the requested distance.
 

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