Archived Centripetal Acceleration and string tension Question

[BlueCardBird]

Homework Statement

A 6.0kg object is attached to two 5.0m-long strings and swung around in a circle at 12 m/s. Determine the tension in the two strings, and explain why the tensions are not the same.

Given:
m=6.0kg
v=12m/s
length of string= 5.0m

Homework Equations

ƩF=m.a
ac=v^2/R

The Attempt at a Solution

There was a diagram that goes with the question but i couldn't simply draw it. Basically its a 8.0m pole with two 5.0 m strings attached to a 6.0kg ball. the strings and the pole form a isosceles triangle. I used the 3,4,5 triangle rule to find the radius which is 3, then ac=v^2/R to find the acceleration which came out to be 48m/s^2. After, used sin law to find angle between ball and radius. When i wrote my ƩF statement one came out to be ƩFy=Tasin37°-Tbsin37°-Fg other statement is ƩFx=Tacos37°+Tbcos37°. Is it possible to substitute these two equations eventhough they're acting in different directions?

 

[QuantumQuest]

The radius found in the OP is right (3 m). The centripetal acceleration is also correct (48 m / s²). The OP uses symbols like Ta, Tb etc. and takes sin law, without giving a diagram, so I'll not comment on that part. I really do not understand, what he/she means by

"Is it possible to substitute these two equations eventhough they're acting in different directions?"Anyway, I'll use a Cartesian coordinate system rather than the sin law used in the OP.
The diagram is shown above.

We take the horizontal and vertical components of the two tensions T₁ and T₂ (see Fig. 2).
We first calculate sin(θ) and cos(θ) [Fig. 1]:
sin(θ) = 3 / 5 and cos(θ) = 4 / 5.

For the vertical components, because there is no motion on the vertical direction, we'll have:
T₁cos(θ) = T₂cos(θ) + mg, so: T₁ - T₂ = mg / cos(θ) = 6.0 x 9.8 / 0.8 $kg \frac{m}{s^{2}}$ = 73.5 N.

For the horizontal components of tension, we see that their sum is the centripetal force for the circular motion of the sphere.
So, T₁ sin(θ) + T₂ sin(θ) = m v² / R , so, T₁ + T₂ = m v² / R sin(θ) = 6.0 x 12² / 3 x 0.6 kgr (m/s)² / m = 480 N.

Now, having T₁ - T₂ and T₁ + T₂ we solve the simple system of equations:
T₁ - T₂ = 73.5 N
T₁ + T₂ = 480 N

We get T₁ = 276.75 N and T₂ = 203.25 N.

We see that the tensions are not equal. This is explained, if we take into account the equilibrium on vertical axis (i.e. the weight of the sphere).