Box Hangs From Rope


by TonkaQD4
Tags: hangs, rope
TonkaQD4
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#1
Oct19-07, 05:44 PM
P: 56
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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Doc Al
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#2
Oct19-07, 05:47 PM
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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.
TonkaQD4
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#3
Oct19-07, 06:12 PM
P: 56
I am still a little confused.
The two forces are Tension and Weight, which if the Acceleration is zero, doesnt this mean that the force of the tension and weight are the same.

TonkaQD4
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#4
Oct19-07, 06:17 PM
P: 56

Box Hangs From Rope


Gravity....

So it would be 61kg times 9.8m/s^2 = 597.8N for part (a)
TonkaQD4
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#5
Oct19-07, 06:21 PM
P: 56
How do I tackle part (b)?
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#6
Oct19-07, 06:31 PM
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Quote Quote by TonkaQD4 View Post
I am still a little confused.
The two forces are Tension and Weight, which if the Acceleration is zero, doesnt 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.

Quote Quote by TonkaQD4 View Post
How do I tackle part (b)?
Using the same method. What's the acceleration in this case?
TonkaQD4
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#7
Oct19-07, 09:35 PM
P: 56
5.10 ???
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#8
Oct20-07, 06:48 AM
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Quote Quote by TonkaQD4 View Post
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?
TonkaQD4
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#9
Oct20-07, 11:07 AM
P: 56
Zero
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#10
Oct20-07, 11:11 AM
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Quote Quote by TonkaQD4 View Post
Zero
Yep. Next!
TonkaQD4
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#11
Oct20-07, 11:26 AM
P: 56
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
TonkaQD4
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#12
Oct20-07, 11:28 AM
P: 56
Part (d) would then be 598N - 311N = 287N

Thanks for your help!!!
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#13
Oct20-07, 11:28 AM
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Perfect! (for b, c, & d!)


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