Can a 84 kg Climber Safely Descend with a Damaged Rope?

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Homework Help Overview

The discussion revolves around a physics problem involving a mountain climber weighing 84 kg who is attempting to descend using a damaged rope that can only withstand a tension of 675 N. Participants are exploring whether the climber can maintain a constant speed without breaking the rope and what the maximum allowable downward acceleration would be.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the forces acting on the climber, including tension in the rope, gravitational force, and potential friction. There are attempts to apply Newton's laws of motion to analyze the situation, with some questioning whether the rope is the only force acting on the climber.

Discussion Status

There is ongoing exploration of the problem, with participants providing insights into the forces involved and questioning the assumptions made about the forces acting on the climber. Some guidance has been offered regarding the equilibrium of forces at constant speed, but there is no explicit consensus on the correct approach or calculations.

Contextual Notes

Participants are working under the constraints of the problem as posed in a homework assignment, which may limit the information available for analysis. There is uncertainty regarding the correct interpretation of forces and how they relate to the climber's acceleration and tension in the rope.

samdiah
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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!
Waiting for your reply
 
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You should know that at a constant speed, all forces are in equilibrium. Take it from there.
 
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.
 
Someone please help!
 
samdiah said:
Someone please help!

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?
 
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?
 
samdiah said:
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?

For constant velocity, the net force equals zero, i.e. the forces are in equilibrium. Is this possible?
 
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!
 
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
Can someone please help me again!
 
  • #11
:

:confused:
 
  • #12
Remember that it's the total force that gives you the acceleration, not just the force from the rope.
 
  • #13
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
srry my bad
 

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