Static Friction and Kinetic Friction Physics Problems With Free Body Diagrams

[@TheOrganicChemistryTutor]

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Thank you man i now understand everything about friction and I can finally do my exam with no worries

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[@kabouch]

As for the missing problem no 4 at 24:26, I would think that it takes no horizontal force to lift up an object vertically since the horizontal acceleration is zero; instead, the net vertical force= Fn+Fp-mg-Fk

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How can I get the full video????

[@senqui]

It's been two years, but I think the question asked to stop the vertical acceleration, not the horizontal one. Edit : The Fy forces are going to be the weight force(mg) and the the static friction(Fs). The normal force is going to be perpendicular towards the wall, and finding the normal force is the same as finding the force required to stop it from falling. So, you use the formula us*Fn = Fs(the max value after which it starts sliding) and then you can rearrange to solve for Fn to get 98N. I think that's the answer, feel free to correct.

[@adilgbadegesin]

I got 100 You can imagine that you are solving a regular friction question(on a horizontal surface) but the catch is the g is unknown so thats what you have to find to. My answer states that if force horizontal on the book is 100N * by the coefficient of static friction 0.3 . 100 * 0.3 = 30 30 counters the 30 that acts down on the book in the first place as the downward force is 30 3 * 10 Keep in mind my gravitational acceleration is 10ms^-2 You'll have to tweak the 9.8 into the formula (Weight * gravitational acceleration) ÷ coefficient of static friction

[@kefikhobragade8191]

@@senqui I have a doubt pls. 1) 21:17 : Why is sumFy =0 but not sumFx=0?

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​@kefikhobragade8191 Because the object is not getting lifted but getting pulled

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[@adilgbadegesin]

I got 100 if g = 10ms But 98 if g = 9.8 You can imagine that you are solving a regular friction question(on a horizontal surface) but the catch is the g is unknown so thats what you have to find to. My answer states that if force horizontal on the book is 100N * by the coefficient of static friction 0.3 . 100 * 0.3 = 30 30 counters the 30 that acts down on the book in the first place as the downward force is 30 3 * 10 Keep in mind my gravitational acceleration is 10ms^-2 You'll have to tweak the 9.8 into the formula (Weight * gravitational acceleration) ÷ coefficient of static friction (3*9.8)/0.3 = 98

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What happens when we pull (to prevent the object from moving) an object which is at rest ? Should the static force be added to the applied force?

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