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Free Fall and Tension in a cable

  • Thread starter swannyboy
  • Start date
  • #1
15
0

Homework Statement



A crate is hanging from a cable as shown in the diagram. A monkey that is inside the crate, sits on a ledge that is 1.225m above the bottom of the crate. The monkey steps off the ledge and when he hits the bottom of the ledge, it takes .125 s to stop.

The tension in the cable is 147 N when the monkey is sitting on the ledge and 49.0N while he is falling.

Calculate the tension in the c able during the .125 s that he is stopping at the bottom of the crate.
[URL]http://img130.imageshack.us/i/52884921.jpg/[/URL]

Homework Equations



Not exactly sure- it's in a booklet for my comprehensive final exam. It seems like it's mostly applications of:
slope = rise/run
Area = 1/2base x height
and F = ma

The Attempt at a Solution



Basically what I've done is this:
Tension in the string = force of gravity

Tension = mg
147N = m(9.8N/kg)
M = 15 kg total

Tension = mg
49.0 = m(9.8N/kg)
m = 5kg

So the monkey's mass is 10 kg.

When the monkey jumps off, he Accelerates at 9.8m/s^2 for 1.225meters

d=1/2vt
2.45m/v = t

A = (v2-v1)/(t2-t1)
A =(V^2)/(2.45m)
24.01 = v^2
V = 4.9m/s

D = 1/2vt
2.45m = (4.9m/s)t
t = 0.5s

I think I'm going in the right direction, but I don't know where to go from here.
 
Last edited by a moderator:

Answers and Replies

  • #2
15
0
Now if I try using momentum..../
F(delta_t) = m (delta_v)
= (10kg)(4.9m/s)
Impulse = 49 kgm/s

49kgm/s = F (0.125s)
F = 392 N

Fnet = fg + fa
Fnet = 49N + 392N
Fnet = 441N

So, is that what the tension would be?
 

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