Can somebody help me find the tension in this string? [closed]

My friends and I are debating over the correct answer for this particular physics problem.

There's a pendulum with a $2kg.$ mass and a $5m.$ string attached to a ceiling at $53^{\circ}$ to the vertical. It's released. At point B, $37^{\circ}$ from the vertical, we are asked to find the tension in the string.

Both of us agree that $v = 2\sqrt{5}$.

My equation was $T\cos 37^{\circ} = 20 + F_c\cos{37^{\circ}}$, which gives me $T = 33N$ I got the upward component of $T$ and set it equal to the weight of the block + the downward component of the centripetal force.

My friend's equation was $T = 20\cos{37^{\circ}} + F_c$ which makes $T = 24N$. She got the component of the weight at $37^{\circ}$.

Conceptually, I think we're both correct, but we get different answers. Could somebody answer who's right and why?

Note: we assume $\cos 37^{\circ} = 0.6, \cos 53^{\circ} = 0.8, g = 10m^2/s$

closed as off-topic by Emilio Pisanty, Qmechanic♦Sep 6 '13 at 23:57

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Firstly, we need to identify the fact that the motion of pendulum about the point of suspension is accelerated circular motion. Therefore, there has to be two forces acting on the pendulum's bob :-
1)Centripetal force radially inwards, responsible for centripetal acceleration
2) tangential force necessary for the tangential acceleration which increases the speed.
In your friend's analysis, the net forces in the radial directions are equated to centripetal force which is correct since the radial acceleration caused by the centripetal force is the only acceleration in that direction which is accounted for in the equation of the centripetal force. But the calculation is wrong and it amounts to $20N$.
But in your analysis, you equated net force in the vertical direction to the vertical component of the centripetal force which is not correct, since centripetal force is not the entire story. This is because, in vertical direction there is a component of tangential acceleration which is not accounted for in your analysis. Therefore, you have to include an additional term for the vertical component of tangential acceleration in the downward direction. Your equation will now be $$T\cos 37=F_c\cos 37 + 20-2a_t\cos 53$$ where $a_t=g\sin 37$ since the only force which can bring about tangential acceleration is the tangential component of weight. This gives $20 N$. In your friend's analysis, the analysis is correct but she probably did some mistake in her calculations which still amount to $20N$. Please correct the calculation and amend your equation then both the methods yield the same result.

The weight at the moment of interest is travelling in a 5 metre radius circle at $2\sqrt5 \, m/s$

So the centripetal force needed is $F_C$:$$F_C=\frac{2\times 20}{5}=8$$

The force of gravity on the object is $20 \, newtons$ The radial component of this force is $20 \times cos(37^\circ)=12 \, Newtons$

The tension must balance this component of gravity and supply an extra 8 Newtons. $T = 20 \, Newtons$