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

  • Thread starter Thread starter indietro
  • Start date Start date
  • Tags Tags
    Work
AI Thread Summary
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.
indietro
Messages
46
Reaction score
0

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!
 
Physics news on Phys.org
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 :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

How Is Work Calculated in a Two-Rope Tension Problem?
Replies
1
Views
5K
Work Done by T1 and T2 in Lowering a Piano
Replies
13
Views
23K
Calculating Work: Lowering a Piano with Two Ropes
Replies
4
Views
9K
How Much Work is Done by Gravity When Lowering a Piano with Two Ropes?
Replies
8
Views
4K
How can net force point in the direction of *negative* work?
Replies
1
Views
4K

Hot Threads

Recent Insights

Back
Top