# Box Hangs From Rope

1. Oct 19, 2007

### TonkaQD4

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

2. Oct 19, 2007

### Staff: Mentor

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.

3. Oct 19, 2007

### TonkaQD4

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.

4. Oct 19, 2007

### TonkaQD4

Gravity....

So it would be 61kg times 9.8m/s^2 = 597.8N for part (a)

5. Oct 19, 2007

### TonkaQD4

How do I tackle part (b)?

6. Oct 19, 2007

### Staff: Mentor

Exactly. The force equation is: T - mg = ma = 0; so T = mg.

Using the same method. What's the acceleration in this case?

7. Oct 19, 2007

### TonkaQD4

5.10 ???

8. Oct 20, 2007

### Staff: Mentor

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?

9. Oct 20, 2007

### TonkaQD4

Zero

10. Oct 20, 2007

### Staff: Mentor

Yep. Next!

11. Oct 20, 2007

### TonkaQD4

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. Oct 20, 2007

### TonkaQD4

Part (d) would then be 598N - 311N = 287N