Tension in Rope for Box at Rest: 0N

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The discussion revolves around calculating the tension in a rope supporting a 61.0 kg box under various conditions. For part (a), when the box is at rest, the tension equals the weight of the box, which is approximately 598N. In part (b), since the box moves upward at a steady speed, the tension remains the same at 598N due to zero acceleration. For part (c), with an upward acceleration of 5.10 m/s², the tension increases to 909N. In part (d), when the box is slowing down with the same acceleration, the tension decreases to 287N.
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A 61.0kg box hangs from a rope. What is the tension in the rope if:

(a) The box is at rest?
(b) The box moves up a steady 5.10m/s?
(c) The box has v_y = 5.10m/s and is speeding up at 5.10m/s^2? The y-axis points upward.
(d) The box has v_y = 5.10m/s and is slowing down at 5.10m/s^2?

This problem seems easy, but because I cannot even get part (a) I am confused and need help.

To me it seems like part (a) should be either 0N or 61N.
F=ma
F=61kg(0m/s^2) ---> box at rest means acceleration is zero right?
F=0N
 
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Analyze the forces acting on the box (there are two) and apply Newton's 2nd law (Fnet=ma). You're right that "ma" = 0, but that means that the net force is zero.
 
I am still a little confused.
The two forces are Tension and Weight, which if the Acceleration is zero, doesn't this mean that the force of the tension and weight are the same.
 
Gravity...

So it would be 61kg times 9.8m/s^2 = 597.8N for part (a)
 
How do I tackle part (b)?
 
TonkaQD4 said:
I am still a little confused.
The two forces are Tension and Weight, which if the Acceleration is zero, doesn't this mean that the force of the tension and weight are the same.
Exactly. The force equation is: T - mg = ma = 0; so T = mg.

TonkaQD4 said:
How do I tackle part (b)?
Using the same method. What's the acceleration in this case?
 
5.10 ?
 
TonkaQD4 said:
5.10 ?
No. 5.10 m/s is the speed, which is steady. (Note that acceleration has units of m/s^2, not m/s.)

So, what's the acceleration of something moving upward at a steady speed?
 
Zero
 
  • #10
TonkaQD4 said:
Zero
Yep. Next!
 
  • #11
Ok so it 598N again.

Now part (c)...
It is accelerating at 5.10m/s^2

T-mg=ma

ma= 61(5.1) = 311N
mg= 598N

T= ma+mg= 311N+598N = 909N
 
  • #12
Part (d) would then be 598N - 311N = 287N

Thanks for your help!
 
  • #13
Perfect! (for b, c, & d!)
 
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