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Static Equilibrium components (pulleys, weights)

  • Thread starter fizzsucks
  • Start date
1. The problem statement, all variables and given/known data

Component diagram:
include all forces, x & y, angles
compare x components (are they equal?)
do y-components equal 0?


2. Relevant equations

Fnet = ma = 0 = F + F + F....
SOHCAHTOA

3. The attempt at a solution

Attached is my diagram with components drawn in. There are two pulleys with a rope strung across them, with a weight on each end, and one tied in the middle. Weight on the left (100g), middle (120g), right (92g). I need to calculate the x and y components, and see how close they add up to equal 0.

Is the y-component for both vectors = (120g)(9.8)? I thought it should be, however if thats the case, the x-components wont be equal and opposite: (.120g)(9.8)tan54 vs (.120g)(9.8)tan50. And how do I account for the other two weights?
 
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kuruman

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Is the y-component for both vectors = (120g)(9.8)? I thought it should be, however if thats the case, the x-components wont be equal and opposite: (.120g)(9.8)tan54 vs (.120g)(9.8)tan50. And how do I account for the other two weights?
The sum of the two y-components is (120g)*9.8.

To account for the other two weights, note that pulleys change the direction of the tension but not its magnitude.
 

kuruman

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Each hanging weight is exerting a force mg on the rope. This force is the tension in that particular rope.
 

kuruman

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It is not correct. Remember, cos = adj/hyp and sin = opp/hyp.
 

kuruman

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It is not a matter of dotting your t's and crossing your i's. The numbers for the components are incorrect because you use sine where you should have used cosine and vice versa.

Also Fnet,y = F1y + F2y - mg, where mg is the hanging weight in the middle. The net force is the sum of all the forces and there are three forces at equilibrium in the middle.
 

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