Mass attached to a rope (simple)

  • Thread starter Thread starter saltyload
  • Start date Start date
  • Tags Tags
    Mass Rope
AI Thread Summary
A 1000 kg mass is lowered by a rope, initially moving at 10 m/s and decelerating at 1 m/s². The tension in the rope must account for both the weight of the mass and the upward acceleration due to deceleration. The correct equation is T = mg + ma, leading to a tension of 11,000 N, not 9,000 N as initially calculated. The key point is that while the mass moves downward, its upward acceleration increases the tension in the rope. Understanding the direction of acceleration is crucial for solving the problem accurately.
saltyload
Messages
8
Reaction score
0
1. A 1000kg mass is lowered by a rope. If the mass is initially moving at 10m/s, and is decelerating at 1m/s2, what is the initial tension T in the rope?



2. T = mg



3. So I tried setting up T = mg, and then since it is decelerating the tension must be less than what it would be if it was in static equilibrium. Then I added ma to side T.

ma + T = mg

Ended up getting T = 9000N, but the answer is 11,000N.


What did I do wrong? I thought my thinking was correct. I thought about a mass in an elevator which would weigh less if elevator moved downward and weigh more if elevator moved up.
 
Physics news on Phys.org
saltyload said:
I thought about a mass in an elevator which would weigh less if elevator moved downward and weigh more if elevator moved up.
If the elevator accelerates downward, the mass would 'weigh' less. Acceleration, not movement, is the key.

So what's the direction of the acceleration of the 1000 kg mass? (Read carefully.)
 
Doc Al said:
If the elevator accelerates downward, the mass would 'weigh' less. Acceleration, not movement, is the key.

So what's the direction of the acceleration of the 1000 kg mass? (Read carefully.)

AH! the mass is moving downward, but since its decelerating, its accelerating upwards, which increases the tension of the rope!
 
Exactly! :wink:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top