Calculating Work: Piano Lowering 5m, 2500N Gravity, T1 1830N, T2 1295N

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The discussion centers on calculating the work done by gravity and two tension forces (T1 and T2) while lowering a piano. The gravitational force acting on the piano is 2500 N, resulting in 12500 J of work done by gravity. T1, with a force of 1830 N, contributes approximately 4575 J of work in the opposite direction. The participant initially miscalculated the work done by T2, which should be around 4579 J, leading to a net work of zero when all forces are considered. The correction highlights the importance of using the correct angle for calculations to ensure accurate results.
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Homework Statement


There is a piano that is being lowered 5 m. It has gravity force acting in y-direction (down) of 2500 N. Then it has two ropes attached, they are vectors that are pointing up and out. so T1 is 1830N [W60\circN] and T2 is 1295N [E45\circN]


Homework Equations


so the work by gravity is 12500 J
the work by T1 is 4575 J (opposite direction to work by gravity)

The Attempt at a Solution


so it would make sense that the work by T2 is somewhere close to 7000 J so that the net work is 0. but when i calculate it i get around 4000 J. Can someone check my math...

For T2:
W = Fcos\theta\Deltar
= 1295*cos45*(5)
= 4579 J

? - thnx!
 
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Hi there,

I believe you have a problem with the angle of T1. Becausem, when I verified your calculations, I come to precisely 0 net work.

Cheers
 
oh yes thank you! i was using an angle that gave the component perpendicular to motion :)
 
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