More on Normal force (shoe on floor) | Physics | Khan Academy

[@gauravGupta-bk2sw]

6:08 whaaaat 🤣 a SHOE!!

[@virupannamedikinal]

Thank you.

[@NylaTheWolf]

Oh my god thank you!!!

[@gautamvashishtha6330]

ah amazing video! its just that i wanted to know that how do i understand or rather obtain the directions of the x and y components of the diagonal force?

[@abhishekpandey9049]

Hi... BTW three years later...

[@gautamvashishtha6330]

@@abhishekpandey9049 😂😂 already got into IIT

[@abhishekpandey9049]

@@gautamvashishtha6330 🎊 congratulations..... Well I hope your pic is the latest one right

[@gautamvashishtha6330]

@@abhishekpandey9049 lol No da that's from a while back. Should update that yeah

[@abhishekpandey9049]

@@gautamvashishtha6330 I'm really sorry to confuse you with the name....but...I'm Parnika actually. 🤗🤗anyway... Hi my name is Parnika ☺☺ (edit) : looking good in that pic.... 😄😄

[@zouraizkhan2467]

when the shoe accelerates toward the center of the earth dont the air particles act like a surface and produce normal force on the shoe

[@pratyushpathania7030]

True but in highschool level you never really consider the air pressure or even air ' existence.

[@Tea-Spin]

If you have a box stay rest on the floor and then pull it up, the momment you pulled it up, does normal force still included in the net force?I mean it's the momment right about when you are going to pull it up so it's still on the floor in the beginning

[@tylerairey1769]

The normal force would decrease to 0 over the time you pulled it up. At any given time from the box being in contact with the ground to being above the ground (not in contact), the normal force would be somewhere between it's original force and 0.

[@saumyadipnandy2060]

What will be the force on the floor if the elevator is accelerating with an acceleration of g and the weight of the shoe is M.

[@tylerairey1769]

I'm guessing you mean if the elevator had an acceleration of g, upwards? Also, in a question you'll generally be given the mass of an object; not the weight, so for simplicity, let's designate the "mass" of the shoe as M. The force downward is m*g, or 9.8M N. The NET force, as you stated in your question, is m*a, or 9.8M N. Therefore, the normal force is 9.8M + 9.8M = 19.6M N. An easy way to think about this is the normal force has to be equal to the force of gravity + the upward force, because it has to overcome gravity to accelerate an object upward.

[@saumyadipnandy2060]

Tyler Airey Hey man thanks for the explanation but I wanted to know the formula for normal force

[@tylerairey1769]

Oh sure, no worries. However, the normal force doesn't have it's own formula, it is just a constituent force along with any other force, that only appears when 2 surfaces are in contact, and contributes to the total net force. Eg. Net force = Fn - Fg (I have assumed here that the normal force, Fn, and the force of gravity, Fg, are acting in opposite directions, with the up direction being positive) I suppose, if you wanted a formula for the normal force, Fn, for these instances, it would be: Fn = Net Force + Fg ... + Fd - Fu (Fu = any additional force(s) "upward") (Fd = any additional force(s) "downward") Be careful to only calculate forces in either the x or y plane, sometimes the normal force (unlike gravity) can be at an angle, a, in which you'd only consider the y component of the normal force, which will often be Fn*Sin(a). Hope this has helped (:

[@alexanderavendano674]

ghostbusters

[@NotLegato]

technically, isn't there always a normal force exerted on particles in the path of a moving object, because nowhere is a perfect vacuum?

[@varda3676]

we call that air resistance(if it's in the air) or viscous drag(in a liquid)

[@anatolykarpov2956]

You didn't fix the equation after saying that a = 0 any more so The last equation is Fn×a = .....