Indeed. My professor didn't teach us the moment arm method, and I have a feeling that if a ladder problem is on my exam today, the distance between the ground and the point where the ladder meets the wall won't be given. Thank you so much for this video. Best I've watched all semester. Wish I found these sooner.

+cstephenmurray hello i am not sure if you mentioned if there is a friction betwen the ladder and the wall or not, because it was not taken into account in the equations I think

Luke Parker Because of something called the "moment arm". Torque is defined as a force acting perpendicular to a lever. When the force is NOT perpendicular to the lever you have two choices: 1) you can resolve the force into its components perpendicular and parallel to the lever or 2) you can find the perpendicular lever, known as the moment arm. Which is better? Not the correct question. Instead: which is easier (physics answer). In the ladder case, way 2 is much easier. How do you find the moment arm? Draw the force infinitely long and then find the distance that is perpendicular to that force. In the video the thick line drawn from the point where the ladder touches the ground is the moment arm. I just so happens that it is the same distance as from the ground to where the wall touches the ladder, which is 5sin v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} b\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} b\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} θ. Below you can see the interior angles. It should be clear, now. 281 7772400 10058400 259 261 257 276 262 279 1 0`````````````````````` 5 1 0 285 282 1 False 0 0 0 0 -1 304800 243 True 128 77 255 3175 3175 70 True True True True True 278 134217728 1 1 -9999996.000000 -9999996.000000 8 Empty 16711680 52479 26367 13421772 16737792 13382502 16777215 Bluebird 22860000 22860000 (`@````````` 266 263 5 110185200 110185200

3.5 m is the distance to Slim Jim from the ground (I just chose that at 1:41). The center of mass of the ladder is at the center of the ladder which is 2.5m.

The wall is frictionless, so there’s no “reaction force” possible there, if you are talking about Newton’s 3rd Law pairs. At the wall the 3rs Law force pair would be “the ladder pushing on the wall” and “the wall pushing back on the ladder”. These are normal forces which are always perpendicular to the surface.

You just mathn't..

Yam Almansour using Pythagoras we know that when the hyp is 5 and one of the other lengths is 3 or 4 then the missing length will be the missing number in this sequence (3,4,5) hence the name 3 4 5 triangle hope this helps

Yam Almansour Pyrthagoras theorem

+anfarahat why would there be? the surface it is in contact with is flat.

+scratch12367 That's a closer model to reality. Friction exists both on the wall and on the floor. I do not see the relation between a surface being flat and frictionless. Can't we have a wall that is flat and rough at the same time?

I don't think it would have any effect on the system unless the floor is frictionless

It does (3rd Law), but we are analyzing the ladder not the Person/ladder system

Sin(theta) = opp/hyp to solve for oppsite it becomes: hyp * sin(theta). And that's what he did. I know it's been 9 months ago but eh.. lol

because 5 is the hyp. 5sin53 is the distance perpendicular to the force.

TrailBlazer65, maybe you didn't search? LOL I have my test in 3 hours and i'm not feeling comfortable with this. :D

In this case what will be the normal reaction applied by floor on ladder?

Well, considering F(friction) = F(normal) * u(coefficient of friction), you can have all of the normal force in the ENTIRE UNIVERSE and it wouldn’t matter, the ladder will just fall “Normally.”

Ladder would fall

Another omission by me, sorry. I SHOULD have started by defining my system. In this case I used the combined system of Slim Jim and the ladder. As a result the normal forces between them are internal forces and can be ignored. If our system was defined as just the ladder, then Slim Jim does apply a normal force to the ladder. Since Slim Jim is also at static equilibrium, so mg = Fn for Jim, then, by Newton's 3rd Law, Fn of Jim on the Ladder also equals mg of Jim. Hope that helps.

@@cstephenmurray Thank you!

Yes, the man is a legend

+Baljeet Singh The frictional force is 100% horizontal. It is counter-acting the force the ladder is exerting horizontally on the wall

wtf?

cstephenmurray lol not at my school