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

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In summary, the problem involves a piano being lowered 5 m with a gravity force of 2500 N acting downwards. Two ropes are attached, with T1 at 1830N [W60\circN] and T2 at 1295N [E45\circN]. The work done by gravity is 12500 J, while the work done by T1 is 4575 J in the opposite direction. It is expected that the work done by T2 will be around 7000 J to achieve a net work of 0, but calculations show it to be 4000 J. After further discussion, it is discovered that the angle used for T1 was incorrect, leading to a net work of
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indietro
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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[tex]\circ[/tex]N] and T2 is 1295N [E45[tex]\circ[/tex]N]


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[tex]\theta[/tex][tex]\Delta[/tex]r
= 1295*cos45*(5)
= 4579 J

? - thnx!
 
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  • #2
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
 
  • #3
oh yes thank you! i was using an angle that gave the component perpendicular to motion :)
 

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

1. How do you calculate the work done in lowering a piano 5m?

The work done in lowering the piano can be calculated using the formula W = Fd, where W is the work done, F is the force applied, and d is the distance moved. In this case, the distance is 5m and the force is 2500N, so the work done is 12,500 joules.

2. What is the role of gravity in this scenario?

Gravity is the force that pulls the piano towards the ground. In this scenario, gravity plays a crucial role in determining the amount of work required to lower the piano. The force of gravity is 2500N, which is equal to the weight of the piano.

3. How do you calculate the tension in the rope at T1?

The tension in the rope can be calculated using the formula T = F + ma, where T is the tension, F is the force applied, and m is the mass of the object. In this case, the force applied is 2500N and the mass of the piano is not given. Therefore, the tension at T1 cannot be accurately calculated.

4. What is the significance of the tension at T2?

The tension at T2 is the force that is required to balance the weight of the piano and keep it from falling. In other words, it is the force that is equal and opposite to the force of gravity. In this scenario, the tension at T2 is 1295N, which means that an additional force of 1295N is needed to keep the piano from falling.

5. Can the work done be negative in this scenario?

Yes, the work done can be negative in this scenario. If the piano is lifted back up to its original height, the work done would be negative since the force applied is in the opposite direction of the displacement. Additionally, if the piano were to fall while being lowered, the work done would also be negative as the force of gravity would be doing the work instead of the person lowering the piano.

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