5.17 | Consider the 52.0-kg mountain climber in Figure 5.20. (a) Find the tension in the rope and
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- Опубліковано 12 вер 2024
- Consider the 52.0-kg mountain climber in Figure 5.20. (a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. (b) What is the minimum coefficient of friction between her shoes and the cliff?
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So thankful for your videos! Having a step-by-step solution is so beneficial, if I get stuck somewhere, or can't find my mistake. Or, even if I have no idea where to start. Really appreciate them, thanks so much!
You're very welcome Valencia! Glad these videos are helpful. Spread the word, and keep working hard! 😀
Dang, I've watched like 3 of your vids now
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Thanx.
But in 16:11 it is the static coefficient of friction not the kinetic !
Coz even when she is stationary (and that is what you built your solution on) the static friction force is the for which is holding the climber from slipping.
That's what I thought because isn't the static friction acting upwards on her shoes, causing them to not slip down?
Extremely helpful, thank you!
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you saved my sanity with this video. Thank you so much :)
You got it Sarah! Happy studying and spread the word! 😀
This helps me a lot thank you so much !
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for part a why would you not include the normal force or the friction force
Woooow this is so great. Thank you
You're welcome, Arshad! I am so glad that these videos are helping you in your Physics class. Keep up the great work and thank you for being a part of this community! 😀
very nice
Keep up the great work! 😀
how come the tension on the rope is more than the tension would be if she were to be hoisted solely by the rope which is ma=(52)(9.8)=509N. isnt the rope losing tension due to her using the wall to support some of her weight rather than the rope gaining more tension? she is stationary, using force on the wall to support her legs, and not being pulled down by anything but gravity. what am i missing?
ty for the vids, helpful as always!!
So, FL is not the normal force on the leg?
this video helped me immensely, this problem was making me want to smash my head against a wall
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this was so helpful thank you
I'm so glad, Fernanda! You're welcome - Keep up the hard work! 😀
im a little bit confused, why is friction kinetic and not static? I thought that friction is static whenever an object/person is stationary because there's no movement
Hi there! Good question - in terms of calculations, using the coefficient of static friction or kinetic friction doesn't matter since you will get the same answer. However, since the question was asking for the "minimum coefficient of friction", the coefficient of kinetic friction was chosen because the coefficient of kinetic friction is ALWAYS lower (minimum) than the coefficient of static friction. If you look at the charts for a system's coefficient of static and kinetic friction, the coefficient of kinetic friction is always less than the coefficient of static friction. Hopefully this helps!
Thank you!!
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@@GlaserTutoring I told my whole physics class to use your channel because it was so helpful. We all wouldn't pass without you!
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Why is F2 Flcos15 and not sin15??
Great question, Aaron! This goes back to the good ol' SOH CAH TOA days of geometry. Finding the X and Y components for ANY right triangle in Physics always obeys SOH CAH TOA. That is why I don't really like to memorize that the x always goes with one trig function and y goes with another trig function, because that is not always the case. In this case we are looking for the the F2x component, so we have to jot down in our mind everything we know in order to solve for Flx. We have to relate it to the overall vector so we have the HYPOTENUSE, we have the ANGLE (15 degrees) and the component is ADJACENT to that angle! So we have an adjacent and hypotenuse. According to SOH CAH TOA, that is COS! That's why we use cos there. Hopefully this helped! We also have a video that goes into depth about SOH COA TOA for Physics if you want to check that out! ua-cam.com/video/z5nkXQz0AV8/v-deo.html