Two Pulley System: Equilibrium & Theta Calculation

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Homework Help Overview

The discussion revolves around a two-pulley system in equilibrium, focusing on the tensions in the ropes and the calculation of angles (theta and phi) related to the system's forces. Participants are exploring the relationships between the weights and tensions in the context of static equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning whether the tensions in the ropes remain constant and how to approach the equilibrium conditions by summing forces. There are discussions about the signs of the weights and whether to consider certain forces as negative. Some participants are also considering the use of right triangles to solve for resultant forces.

Discussion Status

The discussion is active, with participants providing insights into the equilibrium conditions and the relationships between the forces. Some guidance has been offered regarding the need to equate forces to zero, but there is no explicit consensus on the interpretation of the signs of the forces involved.

Contextual Notes

Participants are navigating assumptions about the system, such as the frictionless nature of the pulleys and the need to break down forces into components. There is an ongoing exploration of how to approach the problem without fully resolving the calculations.

Ry122
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Does the tension in the rope from B to A and from A to C remain the same?
To determine the value of theta when the system is in equilibrium do I need to calculate the sum of all acting x and y components? If so would B and C be negative and A be positive?
 
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Ry122 said:
Does the tension in the rope from B to A and from A to C remain the same?

Not in general.

To determine the value of theta when the system is in equilibrium do I need to calculate the sum of all acting x and y components?

At a point which is in equilibrium, you always have to equate the sum of all forces along any direction to zero. The same goes for a system in static equilibrium, but considering all the forces at one time may not always turn out to be so fruitful

In this case, the weight of A is balanced by the resultant of the two tensions, which in turn are equal to the weights of B and C respectively. (The pulleys are assumed to be frictionless, and I've just given you the method of solving the problem.)

If so would B and C be negative and A be positive?

No. (I presume you mean their weights.)
 
if B and C are not negative then what is? something has to be negative otherwise the sum of the system won't be equal to zero.
do i just use a right triangle to solve for the resultant?
 
Last edited:
are B and C positive and A negative?
 
Ry122 said:
if B and C are not negative then what is? something has to be negative otherwise the sum of the system won't be equal to zero.
do i just use a right triangle to solve for the resultant?

There are reaction forces on the pulleys, acting upward. That gives your negative.

I didn't quite understand your 2nd Q. Do you mean how to solve for the forces at where A is hanging from? Break up into horizontal and vertical components, and then equate the sums to zero.
 
does this look right
Bsin(theta)+Csin(phi)-A=0
Bcos(theta)-Ccos(phi)=0
so i don't even have to find the resultant tension i just solve for theta?
 
Ry122 said:
does this look right
Bsin(theta)+Csin(phi)-A=0
Bcos(theta)-Ccos(phi)=0
so i don't even have to find the resultant tension i just solve for theta?

That is correct, but you should show the steps how you got there.
 

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