# Two pulley system

http://users.on.net/~rohanlal/Q40.jpg [Broken]
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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Shooting Star
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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?

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are B and C positive and A negative?

Shooting Star
Homework Helper
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 dont even have to find the resultant tension i just solve for theta?

Shooting Star
Homework Helper
does this look right
Bsin(theta)+Csin(phi)-A=0
Bcos(theta)-Ccos(phi)=0
so i dont 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.