Explains moment arm and torque for a ladder against a wall with friction on the ground.
Пікірлер: 98
@cstephenmurray9 жыл бұрын
OK - here's a link to a document that explains the moment arm for the wall's normal force: www.cstephenmurray.com/Acrobatfiles/aphysics/NotesAndExamples/Rotational/LadderProblemMomentArm.pdf It should make it more clear.
@timm86106 жыл бұрын
I was struggling for hours trying to understand until I watched this video. Thanks a bunch! Add in the distance traveled on the ladder until the ladder starts to slip and this will be gold.
@cstephenmurray9 жыл бұрын
The wall is frictionless. Otherwise it is more difficult. The problem as stated is difficult enough for students as is.
@StoesA10 жыл бұрын
Hello Stephen, Why isn"t there a frictional force (Ffy) preventing the ladder from going down? There is a frictional force,Ff, preventing the ladder from collapsing. Or can the problem be seen as, Ffy and Ff do the same (prevent the ladder from collapsing) and thereby one can be chosen. Would solvingthe problem with Ffy instead of Ff result in the same values? Kind regars
@cstephenmurray9 жыл бұрын
The wall is frictionless. Otherwise it is more difficult. The problem as stated is difficult enough for students as is.
@rashadarbab27694 жыл бұрын
cstephenmurray lol not at my school
@matrixate9 жыл бұрын
Ah yes. The moment arm...that which is most resisted by students. Great video.
@sgtchuckle1177 жыл бұрын
Great video, man. Very useful for finals. However, I learned Torque=rFsin(theta), which made this a bit harder to follow, although the problem solving steps were still helpful.
@FauzanJamain979 жыл бұрын
sir why do you use cos instead of sin.. because from what i learn, we need to use sin because the angle need to be 90 degree...
@draganandrei53564 жыл бұрын
You just mathn't..
@tiraul077 жыл бұрын
Why is r perpendicular for slim jim and the ladder horizontal and for the point at the wall vertical????
@yamalmansour50218 жыл бұрын
can you please explain why did we get 4 m in the beginning the length of the rectangul ?
@MattWoodYT7 жыл бұрын
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
@shantaramchavan5066 жыл бұрын
Yam Almansour Pyrthagoras theorem
@THUNGUNS8 жыл бұрын
Nice tutorial. I wish you were my physics teacher. xD
@njabulongwenya409510 жыл бұрын
thanks it has improve my thinking ability
@TrailBlazer659 жыл бұрын
Why did I not find this earlier than the night before my test!?
@armankhamiszadeh9 жыл бұрын
TrailBlazer65, maybe you didn't search? LOL I have my test in 3 hours and i'm not feeling comfortable with this. :D
@anfarahat9 жыл бұрын
Why there is no upward friction force at the point of contact of the ladder with the wall?
@scratch123678 жыл бұрын
+anfarahat why would there be? the surface it is in contact with is flat.
@anfarahat8 жыл бұрын
+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?
@heinrichnathanielmarlaw4175 жыл бұрын
Very helpful video. Keep Going!
@scratch123678 жыл бұрын
at the very end i think you divided by 4 instead of 5 am i right?
@ezerium88089 жыл бұрын
what if the ladder is weightless ? should i consider it as 0N ?
@mexicobasado81772 жыл бұрын
10:45 he talks about coefficient of friction
@seer92937 жыл бұрын
shouldn't it be 720/361 in the last?
@reddy.chenny_1235 жыл бұрын
dude ur a legend.
@vintageironmotorcycles10 жыл бұрын
Thank you!!
@seer92937 жыл бұрын
btw, Nice video.Thanks!
@XxNinjaLimeXX10 жыл бұрын
Thanks a lot.
@jdvaldez5510 жыл бұрын
Thank you
@dawitawash328510 жыл бұрын
tanks
@jazzm55577 жыл бұрын
Perfecto... this is how my studying works for university physics 1 exams, lol
@jackearhart42655 жыл бұрын
slim jim
@BaljeetSingh-hr5pv8 жыл бұрын
the frictional force should be acting upwards...totally wrong!!
@scratch123678 жыл бұрын
+Baljeet Singh The frictional force is 100% horizontal. It is counter-acting the force the ladder is exerting horizontally on the wall
@Capumaraca8 жыл бұрын
wtf?
@rashadarbab27694 жыл бұрын
12kg = 120N learn something new everyday.
@cstephenmurray10 жыл бұрын
To: IIproductionsII Yeah, when I just rewatched the video I noticed that I could have just used 4m (from the diagram) instead of doing the trig. I was thinking of other things. Either way, this is the part of the process that most of my students struggle with. I think it is important to understand how the moment arm = 4m, then the actual number.
@dudley54249 жыл бұрын
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.
@Dribbles8810 жыл бұрын
Thank you so much, I'm about to go into an exam and this was a great recap for me from start to finish!! You just strengthen my foundation. :D
@21lukeparker9 жыл бұрын
why, at 9.00 did you use cos then cos and then swap to sin? arent you finding out the horizontal components? wont sin find the vertical force component?
@cstephenmurray9 жыл бұрын
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
@INTEGRALPHYSICS3 жыл бұрын
Serious kudos for doing this in PAINT.
@BaliMystic8 жыл бұрын
Great job! I dunno if you mentioned it but I believe you took for granted the force of friction of the wall was negligible
@BetterThanYou967 жыл бұрын
Now, for the same problem (exclude slim Jim), how would I find the magnitude and direction of the force exerted at the bottom of the ladder. Hell, can someone explain to me what tht even means?
@MrCkntobias6 жыл бұрын
Wow, please be my Physics teacher, my current prof is terrible. I literally understood everything here xD
@ayushanand52214 жыл бұрын
Torque not "twerk" lmao
@Deathgravity492 жыл бұрын
Thank you sir 🤙🤙🤙💞💞
@kairoshen31609 жыл бұрын
Can you do a video on a ladder with smooth wall and ground?
@tiborkote8 ай бұрын
In a 3-4-5 triangle its 30-60-90 degree....
@kakakaka-im1pz7 ай бұрын
Hello، I have a few questions you can ask me
@marvinlee765710 ай бұрын
I have a problem, why the wall does not have a vertical upward reaction force (vertical upward friction) in this case? Thank you.
@cstephenmurray10 ай бұрын
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.
@Mount_Currie10 жыл бұрын
Why is the Fwall 5sin53? Shouldn't it be 4sin53? Because the height is 4
@Dribbles8810 жыл бұрын
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
@johndoe-el6ko10 жыл бұрын
because 5 is the hyp. 5sin53 is the distance perpendicular to the force.
@cstephenmurray9 жыл бұрын
OK - sorry everyone. I tried to add a picture that clearly explains the moment arm, but it ended up a bunch of computer code. I will figure that out, add it to the video OR make a new video to explain it.
@saadamiens9 жыл бұрын
+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
@honestvalley9 Жыл бұрын
Thank you. You’ve given my problem-solving-work-sheet marvelous definition.
@farafeu26707 жыл бұрын
I really appreciate the video, but it'd be more helpful if you state right away what are we looking for (ie, the question)
@psilvakimo5 жыл бұрын
No free-body diagram and associated coordinate system. The presentations are haphazard without them.
@chrisbernardmadriaga57406 жыл бұрын
THIS IS A MESS. THE SUMMATION OF MOMENTS PERPENDICULAR DISTANCE MUST BE RESPECTED TO HORIZONTAL AXIS
@Lucky-fc7ld4 жыл бұрын
hahahahaha thoooo.....
@jameshiggins24484 жыл бұрын
Thank you so much. Statics suddenly makes so much more sense.
@yamrajoli38345 жыл бұрын
at 9;06 why u multiplied by 5 instead of 4
@khaleelal-ashhab28716 ай бұрын
Thank you, this was really helpful.
@antaralamin89646 жыл бұрын
Hello, at around 7:20 did you add 2.5 and 3.5 to get 5 meters? please explain because now I'm confused as to if it's wrong or if I missed an important step. Thanks for the helpful video!
@cstephenmurray6 жыл бұрын
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.
@Iamhappy82810 жыл бұрын
Thanks! you made this problem actually make sense to me.
@jamirwesley32972 жыл бұрын
Thank you from Gettysburg Pa
@mathematicianjeff83586 жыл бұрын
Really good video. Thumbs up
@Mount_Currie10 жыл бұрын
nevermind, I figured it out...
@abenagyampo48452 жыл бұрын
I'm struggling to understand why there is no normla force of the ladder on the person, or if there is why it's not included in the free body and any of the calculations?
@cstephenmurray2 жыл бұрын
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.
@abenagyampo48452 жыл бұрын
@@cstephenmurray Thank you!
@ruthferez45213 жыл бұрын
excellent explanation thanks
@akauth79 жыл бұрын
Great video. Very clear and helpful.
@scratch123678 жыл бұрын
Thanks you helped me out
@claramelb74765 жыл бұрын
Thank you! It actually makes sense
@azwindinikhathutshelo31917 жыл бұрын
Really useful :-}
@amarsoni35786 жыл бұрын
Thank you so much
@timothykalio1576 жыл бұрын
nice and thick
@anteater25369 жыл бұрын
thank you sir.
@FlaminKokeVCX5 жыл бұрын
Why does the ladder not exert a normal force on the person?
@cstephenmurray5 жыл бұрын
It does (3rd Law), but we are analyzing the ladder not the Person/ladder system
@zarinh92406 жыл бұрын
G
@indrejithjayaprakash97249 жыл бұрын
Thanks a lot :)
@MrBomb728 жыл бұрын
Thank you for this!!!
@umarfaruq16377 жыл бұрын
slim jim is travelled around the world...
@timm86106 жыл бұрын
Yes, the man is a legend
@ishikasingla95226 жыл бұрын
Sir what will be the situation if floor and wall...both are frictionless??
@ishikasingla95226 жыл бұрын
In this case what will be the normal reaction applied by floor on ladder?
@charlesheilweil87296 жыл бұрын
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.”
@timm86106 жыл бұрын
Ladder would fall
@tirthadas57676 жыл бұрын
What if the wall is not frictionless??
@timm86106 жыл бұрын
I don't think it would have any effect on the system unless the floor is frictionless