Calculate Leg Breaking Force: 60kg, 1m Height Drop

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Homework Help Overview

The discussion revolves around calculating the force experienced by the body when landing from a height of 1 meter, specifically focusing on the implications for leg injuries. The subject area includes concepts from mechanics and material science, particularly relating to force, stress, and impulse.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants express confusion regarding the stopping distance and the implications of landing stiff-legged. Some are questioning the lack of clear formulas applicable to the problem, while others are exploring the concept of impulse in relation to the stopping distance.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the problem's phrasing and the underlying physics concepts. There is a mix of confusion and attempts to relate the problem to known physics principles, but no consensus has been reached on how to approach the calculations.

Contextual Notes

Some participants note the potential for misunderstanding due to the phrasing of the problem, particularly regarding the concept of stopping distance during a fall. There is also a mention of imposed homework rules, such as avoiding direct solutions.

Faiza
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If you jump from a desk and land stiff-legged on a concrete floor,
the chance is good that you will break a leg. To see how this
happens, consider the average force stopping your body when you
drop from rest from a height of 1.00 m and stop in a much shorter
distance d. Your leg is likely to break where the cross-sectional area
of the bone (the tibia) is smallest. This is at a point just above the
ankle, where the cross-sectional area of the bone is about 1.60 cm^2.
A bone fractures when the compressive stress on it exceeds about
1.60*10^8 N/m^2. If you land on both legs, the maximum force your
ankles can safely exert on the rest of your body is then about……..

2 (1.60*10^8 N/m^2) (1.60*10^-4 m^2) = 5.12*10^4 N.

Calculate the minimum stopping distance d that does not result
in a broken leg if your mass is 60.0 kg. Don’t try it! Bend your knees!
 
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? :'( anyone know how to solve this problem?
 
there arnt any formulas to apply...:[
 
Helllllllllllllpppppppppppppppppppppp
 
I think bumping is illegal, but who knows. I read through this and the phrase "...stop in a much shorter distance d" is confusing me. If you are falling, why would there be a stopping distance?
 
e(ho0n3 said:
I think bumping is illegal, but who knows. I read through this and the phrase "...stop in a much shorter distance d" is confusing me. If you are falling, why would there be a stopping distance?

The stopping distance is the distance your body traveled (very very small) while comnig to a rest. To faiza, think of the problem in terms of impulse: F*deltaS = impulse. A really large impulse is generated by a karate chop as opposed to a punch by a boxer because of the time it took for the force being applied to be removed (small in karate chop's case and large in the boxer's case). It's the impulse that hurts not the force.
 
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