Tension+ centripetal force

Two wires are tied to the 400 g sphere shown in figure. The sphere revolves in a horizontal circle at a constant speed of 8.30 m/s

1) What is the tension in the upper wire?
2) What is the tension in the lower wire?

okay i know that m=.4 and v=8.3
i dont know r, which i thought that maybe i could find by doing the square root (1^2-.5^2) =.866
and then Force= m V^2 / R sooo.. (.4)(8.3^2)/(.866) = 31.819 . but i dont know where to go from there?

okay i also think..
Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
Fnety= Tsin 26 + -T2 sin 26 -Fg= mv^2/ r

so my free body diagram looks like 1 wire pulling in the 2nd quad, one in the 3 quad, and Fg going straight down on the y axis

yep. No motion in y-dir., so forces sum up to zero. In X- dir. they will contribute to centripetal force.

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but i dont understand how the tension in the 2 wires are going to be different

Fnety= Tsin 26 + -T2 sin 26 -Fg= 0
T1 sin 26 - T2 sin 26 = Fg
T1 sin 26 - T2 sin 26 = 3.92

Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
-T1 cos 26 + -T2 cos 26 = 31.82

^ i am not sure how to solve for T1 or T2 . can you please help me with that

i figured out the T2, but now i cant figure out T1. shouldnt it just be plug T2 into one of the equations and it should give you T1??

Doc Al
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Fnety= Tsin 26 + -T2 sin 26 -Fg= 0
T1 sin 26 - T2 sin 26 = Fg
T1 sin 26 - T2 sin 26 = 3.92
Where does the 26 degrees come from?

Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
-T1 cos 26 + -T2 cos 26 = 31.82
Which way does the centripetal acceleration act?

Once you correct your equations, isolate the T1 term in each, like this:

$$T_1 \sin\theta = mg + T_2 \sin\theta$$

26 came from inverse tan (.5/1)

does centripetal acc act on the Horizontal force?

Doc Al
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26 came from inverse tan (.5/1)
But the 1 m side is the hypotenuse. That should be inverse sin (.5/1).

Doc Al
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does centripetal acc act on the Horizontal force?
My question is: What's the direction of the centripetal acceleration?

umm this may be wrong, but i thought centripetal acc just means that the direction is toward the center

ohh wait i know what i did. it is just the angle that is messing me up! thank you for pointing that out. i dont know why, but the angle i used still let me get the tension of the first string??!

Doc Al
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umm this may be wrong, but i thought centripetal acc just means that the direction is toward the center
That's exactly right. So, what direction is that? Should it be positive or negative?

negative

would that be right?

Doc Al
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negative
Good. So fix your equation for horizontal forces.

Fnetx= -T1 cos 26 + -T2 cos 26 = - mv^2/ r

Doc Al
Mentor
Fnetx= -T1 cos 26 + -T2 cos 26 = - mv^2/ r
Good. Now combine with the other equation and solve.