# Homework Help: Falling Mountain Climber

1. Oct 9, 2006

### samdiah

Hello,
I cannot solve this question from 4u physics Dynamic assignment and was wondering if somesome can give me guidance as to how to approach the question?

84 kg mountain climber rapples down a rope. The rope is damages and can withstand only 675 N of tension. Can the climber limit the descent to a constant speed without breaking the rope? If not, what value can the climber limit the downward acceleration to?

Thank You!

2. Oct 9, 2006

### drpizza

You should know that at a constant speed, all forces are in equilibrium. Take it from there.

3. Oct 9, 2006

### samdiah

Thanks for the reply, but I still don't understand.
This is how far I got.
I said FT=675 N
m=92 kg

F=ma
675=92a
a=7.3 m/s2

so the in order for the rope to not break the climber has to travel at minimum acceleration of 7.3 m/sec2. In order for him to attain a constant veloctiy he will have to put more tension on the rope and if he does that the rope will break.

4. Oct 9, 2006

### samdiah

5. Oct 9, 2006

Newton's law of motion states that the resultant (i.e. the sum) of all forces acting on a body equals, in a special case, the product of mass and acceleration, F=ma. Is the force of the rope the only force acting on the climber?

6. Oct 9, 2006

### samdiah

Well there would be gravity and friction present. So should I add
g=9.8*92
=901.6 N
Fnet=226.6 N (down)
a=2.5 m/s (down)

How does that help calculate weather constant velocity is possible?

7. Oct 9, 2006

For constant velocity, the net force equals zero, i.e. the forces are in equilibrium. Is this possible?

8. Oct 9, 2006

### samdiah

Thanks a lot for all that help! This was very helpful for me and saved me a lot of time of trying to understand the question. This is an awesome site with awesome people!

9. Oct 10, 2006

### samdiah

I had another quick question aboutthe second part: what value can the climber limit the downward acceleration to?

is what I did before right:
g=9.8*92
=901.6 N
Fnet=226.6 N (down)
a=2.5 m/s (down)

or do I have to do the following:
F=ma
675=92a
a=7.3 m/s2

Can u explain which one and why? I am kind of unsure.

10. Oct 10, 2006

### samdiah

11. Oct 11, 2006

### samdiah

:surprised :

12. Oct 11, 2006

### Jim Lund

Remember that it's the total force that gives you the acceleration, not just the force from the rope.

13. Oct 11, 2006

### feod2003

I Need Major Help Fast, Its About Finding Time!

How would you get time for this:

5. A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the ground? (Use GUESS Method)

G: vi = 8.0 m/s
g = 9.8 m/s2
dy = 78.4 m
U: dx
E: d = vi*t +1/2 a*t^2
S: dx = 32 m

vi =initial velocity
^2= squared
*= times

14. Oct 11, 2006