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 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.
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.
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.
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.
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
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?
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
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?
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?
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.
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.
+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
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!
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.
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
+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?
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.”
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.
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.
The wall is frictionless. Otherwise it is more difficult. The problem as stated is difficult enough for students as is.
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
Serious kudos for doing this in PAINT.
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.
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.
Great job! I dunno if you mentioned it but I believe you took for granted the force of friction of the wall was negligible
Thank you from Gettysburg Pa
Thank you, this was really helpful.
Wow, please be my Physics teacher, my current prof is terrible. I literally understood everything here xD
Ah yes. The moment arm...that which is most resisted by students. Great video.
Thanks! you made this problem actually make sense to me.
Great video. Very clear and helpful.
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.
Really good video. Thumbs up
thanks it has improve my thinking ability
I really appreciate the video, but it'd be more helpful if you state right away what are we looking for (ie, the question)
Very helpful video. Keep Going!
Can you do a video on a ladder with smooth wall and ground?
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
Nice tutorial. I wish you were my physics teacher. xD
excellent explanation thanks
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?
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
Thank you! It actually makes sense
Why did I not find this earlier than the night before my test!?
TrailBlazer65, maybe you didn't search? LOL
I have my test in 3 hours and i'm not feeling comfortable with this. :D
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?
slim jim is travelled around the world...
Yes, the man is a legend
dude ur a legend.
Thank you sir 🤙🤙🤙💞💞
Why is r perpendicular for slim jim and the ladder horizontal and for the point at the wall vertical????
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?
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!
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.
+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
No free-body diagram and associated coordinate system. The presentations are haphazard without them.
Thanks you helped me out
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!
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.
10:45 he talks about coefficient of friction
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...
You just mathn't..
I have a problem, why the wall does not have a vertical upward reaction force (vertical upward friction) in this case? Thank you.
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 for this!!!
Why is the Fwall 5sin53? Shouldn't it be 4sin53? Because the height is 4
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.
can you please explain why did we get 4 m in the beginning the length of the rectangul ?
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
In a 3-4-5 triangle its 30-60-90 degree....
Thank you so much
shouldn't it be 720/361 in the last?
what if the ladder is weightless ? should i consider it as 0N ?
at 9;06 why u multiplied by 5 instead of 4
Why there is no upward friction force at the point of contact of the ladder with the wall?
+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?
Where is the question
at the very end i think you divided by 4 instead of 5 am i right?
Thank you
btw, Nice video.Thanks!
Thanks a lot.
Why does the ladder not exert a normal force on the person?
It does (3rd Law), but we are analyzing the ladder not the Person/ladder system
Sir what will be the situation if floor and wall...both are frictionless??
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
Hello، I have a few questions you can ask me
thank you sir.
What if the wall is not frictionless??
I don't think it would have any effect on the system unless the floor is frictionless
Thank you!!
Perfecto... this is how my studying works for university physics 1 exams, lol
Really useful :-}
tanks
nevermind, I figured it out...
Thanks a lot :)
Torque not "twerk" lmao
nice and thick
Hi
THIS IS A MESS. THE SUMMATION OF MOMENTS PERPENDICULAR DISTANCE MUST BE RESPECTED TO HORIZONTAL AXIS
the frictional force should be acting upwards...totally wrong!!
+Baljeet Singh The frictional force is 100% horizontal. It is counter-acting the force the ladder is exerting horizontally on the wall
wtf?
12kg = 120N learn something new everyday.
slim jim
hahahahaha thoooo.....
G
The wall is frictionless. Otherwise it is more difficult. The problem as stated is difficult enough for students as is.
cstephenmurray lol not at my school