Solving PSE6 6.P.011: Find Tension in Strings

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The discussion revolves around solving a physics problem involving a 3.85 kg object attached to a vertical rod by two strings while rotating at a constant speed of 7.30 m/s. Participants express frustration over finding the correct force equations to determine the tensions in the upper and lower strings. A suggested approach includes using the equations T"sin(alpha) - T'sin(alpha) - mg = 0 and T"cos(alpha) + T'cos(alpha) = mv^2/r, where T" and T' represent the tensions in the upper and lower strings, respectively. Additionally, users are encouraged to refer to their sketches to find the values of sin(alpha) and cos(alpha). The thread highlights the common difficulties faced in applying centripetal acceleration and force equations in this context.
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Everyone I know is having problems with this question:

3. [PSE6 6.P.011.] A 3.85 kg object is attached to a vertical rod by two strings as in Figure P6.11. The object rotates in a horizontal circle at constant speed 7.30 m/s.

Figure P6.11

(a) Find the tension in the upper string.
N
(b) Find the tension in the lower string.
N

http://www.webassign.net/pse/p6-11.gif for the figure

I've tried using various force equations, only to be wrong. Does anybody know how to do this problem??
 
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Anybody know how to do this problem? Tried a lot of things, but I'm down to my last chance to answer the question correctly
 
it seems you are desparate, just show me your "force equation and I help you out right know
 
Well I tried some forms of the centripetal acceleration formula for the upper tension, and I also tried 3.85*9.88(cos of the angle) + 1/2 the horizontal force, but they haven't worked out.
 
Only because you try to do something, I'm going to help.
Use these forces ==>
T"sin(alpha)-T'sin(alpha)-mg=0 and T"cos(alpha)+T'cos(alpha)=mv^2/r
where T" is the upper tension and T' is the lower tension. Also, if you want to find
cos(alpha) and sin(alpha) use your sketch
 
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