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.

I CRIED TWICE ON MY PHYSICS HOMEWORK AND YOUR VIDEO SAVED MY LIFE

@antaralamin8964

6 жыл бұрын

I've been crying since the semester started :,(

@morganmarshall3605

5 жыл бұрын

you are not alone!!

Thank you. You’ve given my problem-solving-work-sheet marvelous definition.

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.

Thank you so much. Statics suddenly makes so much more sense.

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

The wall is frictionless. Otherwise it is more difficult. The problem as stated is difficult enough for students as is.

Thanks! you made this problem actually make sense to me.

Great video. Very clear and helpful.

thanks it has improve my thinking ability

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

Thank you for this!!!

Thank you, this was really helpful.

Thank you! It actually makes sense

Great job! I dunno if you mentioned it but I believe you took for granted the force of friction of the wall was negligible

Can you do a video on a ladder with smooth wall and ground?

Thank you from Gettysburg Pa

Serious kudos for doing this in PAINT.

Very helpful video. Keep Going!

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?

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.

@dudley5424

9 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.

Really good video. Thumbs up

Why is r perpendicular for slim jim and the ladder horizontal and for the point at the wall vertical????

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.

Thanks you helped me out

Ah yes. The moment arm...that which is most resisted by students. Great video.

what if the ladder is weightless ? should i consider it as 0N ?

excellent explanation thanks

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!

@cstephenmurray

6 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.

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?

@cstephenmurray

9 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

dude ur a legend.

Thank you!!

I have a problem, why the wall does not have a vertical upward reaction force (vertical upward friction) in this case? Thank you.

@cstephenmurray

6 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.

Thank you so much

Wow, please be my Physics teacher, my current prof is terrible. I literally understood everything here xD

thank you sir.

Thanks a lot.

Thank you

I really appreciate the video, but it'd be more helpful if you state right away what are we looking for (ie, the question)

shouldn't it be 720/361 in the last?

can you please explain why did we get 4 m in the beginning the length of the rectangul ?

@MattWoodYT

7 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

@shantaramchavan506

6 жыл бұрын

Yam Almansour Pyrthagoras theorem

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...

@draganandrei5356

4 жыл бұрын

You just mathn't..

at the very end i think you divided by 4 instead of 5 am i right?

btw, Nice video.Thanks!

Nice tutorial. I wish you were my physics teacher. xD

Thanks a lot :)

Why there is no upward friction force at the point of contact of the ladder with the wall?

@scratch12367

8 жыл бұрын

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

@anfarahat

8 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?

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.

@saadamiens

8 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

Thank you sir 🤙🤙🤙💞💞

Why is the Fwall 5sin53? Shouldn't it be 4sin53? Because the height is 4

@Dribbles88

9 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-el6ko

9 жыл бұрын

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

at 9;06 why u multiplied by 5 instead of 4

Sir what will be the situation if floor and wall...both are frictionless??

@ishikasingla9522

6 жыл бұрын

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

@charlesheilweil8729

6 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.”

@timm8610

6 жыл бұрын

Ladder would fall

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?

@cstephenmurray

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.

@abenagyampo4845

Жыл бұрын

@@cstephenmurray Thank you!

Hello، I have a few questions you can ask me

Why did I not find this earlier than the night before my test!?

@armankhamiszadeh

9 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

What if the wall is not frictionless??

@timm8610

6 жыл бұрын

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

Why does the ladder not exert a normal force on the person?

@cstephenmurray

5 жыл бұрын

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

Really useful :-}

tanks

slim jim is travelled around the world...

@timm8610

6 жыл бұрын

Yes, the man is a legend

nice and thick

Perfecto... this is how my studying works for university physics 1 exams, lol

In a 3-4-5 triangle its 30-60-90 degree....

No free-body diagram and associated coordinate system. The presentations are haphazard without them.

10:45 he talks about coefficient of friction

nevermind, I figured it out...

G

slim jim

THIS IS A MESS. THE SUMMATION OF MOMENTS PERPENDICULAR DISTANCE MUST BE RESPECTED TO HORIZONTAL AXIS

Torque not "twerk" lmao

the frictional force should be acting upwards...totally wrong!!

@scratch12367

8 жыл бұрын

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

@Capumaraca

7 жыл бұрын

wtf?

12kg = 120N learn something new everyday.

hahahahaha thoooo.....

The wall is frictionless. Otherwise it is more difficult. The problem as stated is difficult enough for students as is.

@rashadarbab2769

3 жыл бұрын

cstephenmurray lol not at my school