Tension and Pulley Madness (Load weight)

AI Thread Summary
The discussion centers on the calculation of tension in a pulley system involving a load P and various angles. The participant initially assumes the tension is 850 Newtons based on the constant tension principle across the rope but realizes this is incorrect. They clarify that there are multiple ropes with different tensions affecting the system, leading to a reevaluation of their equations. After correcting their approach, they derive the tension T as 1161.2N and the load P as 2322.24N. The conversation concludes with a query about measuring tension in a catapult for a science project.
jegues
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Homework Statement


See figure.


Homework Equations


Sum of Fy = 0
Sum of Fx = 0
Tension is constant across the cord, regardless of the number of pulleys.

The Attempt at a Solution



My first thought after looking at the B A E connection to load P was that the hypotenuse of the 60 degree triangle must be 850 Newtons since tension is constant across the rope. Is there any flaws in my reasoning? (I can't seem to get the correct answer so there must be )



Also doesn't the tension in the cord ABCD depend on load P as well? The heaver P is, the greather the tension in the cord will have to be to keep the cord-pulley system from moving. I'm not sure what forces I'm suppose to use to obtain a tension across ABCD.



Any help is greatly appreciated.
 

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Hi jegues! :smile:
jegues said:
My first thought after looking at the B A E connection to load P was that the hypotenuse of the 60 degree triangle must be 850 Newtons since tension is constant across the rope. Is there any flaws in my reasoning? (I can't seem to get the correct answer so there must be )

There are two ropes between A and B.

Yes, 850 N is the tension in one rope.

But you still have a tension T in the other rope (along AB) (this is like that other thread of yours), and the same tension T along the rope BC.

Try again. :smile:
 
Here's my attempt at a FBD, is this correct?

EDIT: AHA! Got it ;)

F_{x} :

-(850 + T)cos60 + Tcos30 = 0

T = 1161.2N

F_{y} :

(850 + T)sin60 + Tsin30 - P = 0

P = 2322.24N
 

Attachments

  • FBDPM.JPG
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Last edited:
Excellent! :biggrin:
 
how do you measure tension on a catapult because I'm doing a science project and i don't know how
 
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