Force when slipping and falling

[ljlugg]

Help me find the friction force when slipping/falling

I already posted this question, but I needed to reword it and include more info.

A 158 lb man runs on a horizontal wet surface at 3 mph before decelerating to 1.5 mph about 4 feet before a right turn. There are cement stairs at the turn that he plans on using to enter a home. Before he gets to the turn he slips (at this point his right foot has made it onto the cement, whereas his left foot is still on the grass). As he slips he is facing forward on the horizontal surface and his body weight shifts erratically forward onto his right side...which results in a fractured lower right fibula and ankle. Let’s assume that the fractural load level for this person to sustain a foot-ankle fracture is 11 N. The injury resulted in a closed, spiral fracture.
(1) List and estimate the forces involved.
(2) Also list the force involved when compared to the injury tolerance of 11N.

I already figured some things out on my own, such as the fracture load levels etc. I can actually place more assumptions into this story if I need to, so if more info is needed to answer questions 1-2 the experts answering my questions are free to place in additional assumptions at their will, in order to complete my story. I'm doing a biomechanical reconstruction of an accident of my choosing. The only things that are known are the things that I've already placed into the story, I more can be added so long as it doesn't conflict with the data I already presented in the story.
But here's where I'm stuck: I'm stuck on the part where I need to find the friction force involved, in order to answer questions 1-2. I copied/pasted the instructions on how to find the friction force below. I hope they apply to my question though. I need to know if they don't apply. I also need to know if the information I have is good enough to anser questions 1-2 and if not can the experts add their own assumptions into the story?

To find the friction force you need to first find the coefficient of friction, which is equal to tan(θ), where θ is the angle from the horizontal where an object placed on top of another starts to move. For a flat surface, you can pull an object across the surface with a Newtonmeter attached. Divide the Newtons required to move the object by the object’s mass to get the coefficient of friction.

Is this the right formula to find the friction force? Can someone give me a better formula because this one is really confusing?

 

[jvicens]

Thank you for your question. I can help you determine the friction force involved in this scenario.

Based on the information provided, we can estimate the forces involved as follows:

1. Weight force: The man weighs 158 lbs, which is equivalent to approximately 711 N.

2. Friction force: This is the force that resists the motion of the man's body as he slips and falls. To calculate this force, we need to first find the coefficient of friction between the man's shoes and the wet surface. The formula you have provided is correct, but it may not be applicable in this scenario since the man is not sliding on a flat surface. However, we can make some assumptions and use a simplified formula to estimate the friction force.

Let's assume that the coefficient of friction between the man's shoes and the wet surface is 0.5. This means that the friction force is equal to 0.5 times the weight force, which is approximately 355 N.

3. Deceleration force: As the man slows down from 3 mph to 1.5 mph, there is a deceleration force acting on his body. This force is equal to the mass of the man (71 kg) multiplied by the change in velocity (1.5 mph or 0.67 m/s), which results in a force of approximately 48 N.

4. Impact force: When the man's body hits the ground, there is an impact force acting on his lower right leg. This force is difficult to estimate without knowing the exact circumstances of the fall, but we can assume that it is greater than the injury tolerance level of 11 N.

Now, comparing these forces to the injury tolerance level, we can see that the impact force is the most significant and likely responsible for the fracture. The friction force and deceleration force may have contributed to the fall, but they are not strong enough to cause a fracture on their own.

To answer your second question, the friction force of 355 N is significantly higher than the injury tolerance level of 11 N, which further supports the idea that the fracture was caused by the impact force rather than the friction force.

In conclusion, the estimated forces involved in this scenario are the weight force of 711 N, the friction force of 355 N, the deceleration force of 48 N, and the impact force (unknown). Based on these estimates, it