Question

Challenging question: while climbing a chute, what force must you apply to against the walls to support your weight? Coefficient of friction b/w wall-boots=0.56 and wall-gloves=0.39. assume you exert equal forces with hands and feet.

Answer 1

i got an answer in terms of M but i think the question asks for a solid answer

Answer 2

I find it weird that they didn't give you a specific mass to deal with at least

Answer 3

i know. thats uni bro

Answer 4

i used torque but i still arrived at the same answer

Answer 5

oh, srry then I can't help much I am doing gr.11 physics :(

Answer 6

oh good luck! its a grade 10 question tho

Answer 7

Assume that you are being asked to derive a formula that when given a mass (different people), the required force can be determined relative to that mass. Then the formula (and question) would not require a given mass.
Note that the question states "while climbing" so you may need to provide a force that is applied to only three of the four extremities.
This might be a trick question but two conditions would exist while climbing: 1) one hand is being moved so that the total force needs to be applied to two boots and one glove; 2) one foot is being moved so that the total force needs to be applied to one boot and two gloves.
Also, lets assume one boot + one glove, two boots only, or two gloves only are not practical conditions.

Answer 8

hmmm i'm in 7th grade

Answer 9

It's a tricky question. Here is my attempt:

What happens if the climber tries to climb in an inverted position (their hands at the bottom and their feet at the top? My first thought was that, logically, the climber will have to exert a force with their hands to avoid going straight down, but almost certainly their legs will perform a circular motion out of the wall, making them fall anyway.



And so my conclusion is that when climbing, your feet need to counteract your weight and your hands must provide a force perpendicular to the wall counteracting the torque that goes outward the wall.



Basically, your feet act as a pivot point. If you think about it, when you start moving your legs upwards but keeping your hands in a fixed position in the wall, you would have to grab with a greater force. It’s consistent with the fact that by moving your legs, you are decreasing the distance from the pivot point, and hence requiring a greater force to overcome the torque.

Another similar example is one of a car toy. If you put it on a book making an incline, it will go downward. If you put your finger on its rear, you are preventing it from doing so. If you make the angle of the incline equal to 90º, your finger will support its weight but certainly it will perform a circular motion out of the book’s surface.

You’re told the \(F_{feet} = F_{hands}\). Both forces must equal their respective force of friction, otherwise there will be a sliding straight down, a torque, or both. They both require a Normal Force, and that of the hands will have to be provided entirely by the grab. And this force, as far I understand, is the only force against the wall and the force you're asked to find.

Under my assumptions:

\(\large{F_{feet} = w}\)

\(\large{F_{feet} = F_{hands}}\)

\(\large{w = F_{hands}}\)

\(\large{w= \mu_{gloves}N}\)

\(\large{N = \frac{w}{\mu_{gloves}}}\)

Maybe you already have the correct answer. It would be great if you can share it here (: