Physics 15 Torque Example 7 (7 of 7) The Ladder Problem (should be cos(15) at end)
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- čas přidán 28. 08. 2024
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In this seventh of the seven part series we will find out a) Will the ladder slip when the person is 1/3 way up the ladder? and b) Can the person climb safely to the top of the ladder?
I doubt you'll see this after 9 years, but you just SAVED me for my AP physics C exam tomorrow. I've been missing class a lot for some extracurriculars, and have not been grasping the concepts, and this really caught me up. Thanks
Glad you found our videos and good luck on your exam tomorrow! 🙂
Learned more in 14:36 min than I did in my 1 hour and 40 min college lecture class.. thank you!!
im taking a physics 1 in summer rn and im learning the same material u did 7 years ago,i just want to ask which major u did and how is ur life[ur top comment btw]
Gareth.
You are absolutely correct.
I tell my students that one can use either convention as long as one stays consistent. It is when you use vector quantities, it is better to stick with clockwise = negative torque.
You made my physics journey brave!
I fallened a approximate physics teacher once, and passed 1 year, thats too big.
but likely i discovered you, prettiest moment ever in my life.
Now i dream through physics with confident eyes.
Salute you sir!💟
That's awesome!
Our physics teacher is so horrible that she had us watch this in class after giving up trying to teach it. Praise be to Biezen!
you're a really good teacher ! thanks to you I aced my physics exam :)
Congratulations!
Thank you for your clarity in teaching. I wish you are my physics teacher...
Very much admire the shortcuts, canceling of factors, applying fractions to represent ladder’s length, and using the two words ‘dry’ and ‘solid’ in your conclusion. Excellent.
Glad it was helpful!
Francisco,
You are correct
--*10/10*--
I understood everything about torque in the length of this video than how much time I spent in my Physics class.
Bravo for your clarity of explanation on how to pinpoint torques in problems.
I noticed that the Denominator has the Cos 30 ... I also noticed that others have pointed that out... I also Know that Michel is aware of these things... but what surprised me is that when I redid the calculation using the Cos 15 instead of Cos 30 degrees.. the answer does come out to 249.5 N ... in other words.. Michel DID type in Cos 15 when he used his calculator..... Way to go MICHEL!! excellent video.....
Philip,Thanks, that is good to know. Sometimes I look back and wonder: "how did I make such a mistake", but that is part of being human.
well... now you know why your Better Half has Job security in regards to Future BLOOPER REELS.. LOL.... thanks Michel.... laughing here.... it's all good!!
THANKS SO MUCH!!! Finals would have been hell without this channel
This good sir explained this better than poor handwriting on Chegg, hats off to you!
Glad you think so!
you're always on top of these physic problems.
thank you
who dislike this video are .., I think they don't know About the mechanics..Such wonderful explanation of the problem ..Your very well . if Like you a person may as my faculty.. I will be in top position in my college.. Thanks to you Sir ..For such a clear explanation about ladder problems ..
i had a great statics teacher in college at URI in 1975. I took the book apart and literally did every problem. The old slip versus slide problems I could do in my head. Now many decades later, I get it, but would struggle a bit to set up the statically determinate equations, Sum F = 0, Sum Moments = 0
Your channel is very useful ...now I can solve the hard problems that my teacher challenges me to solve without worry!!!Thanks for this detail explanation although I don't understand the language very well ...
Siduduzo
1/3 is correct
you are the greatest physics teacher ever!!!!!!!!!!!!!
Michel, I enjoy your lectures enormously. If only such guidance had been readily accessible when I was at school,
it would have changed my life.
I eventually became a Civil Engineer and part-time college lecturer.
A few little crits:
Draw a good, clear, even exaggerated, diagram;
Keep your "sums" flowing down the board in one direction, even if it means a rolling board. "Workings out" well to the RHS.
Dump the distracting calculator and just give the answer.
Keep up your inspirational work!
Joe
Thank you for your input.
A bit confused with torque. Why when summing the torque's about the base of the ladder do you not employ the angle of the ladder (mg*cos(theta))? For example, as the ladder approaches an angle of 90 degrees w/respect to the horizontal, the climber yields no significant torque as it is nearly || to the lever arm. Thanks.
Physics I is coming to an end and I still didn't understand torque until i watched your videos. Thank you!
Eurekka! I got the best teacher online! Ty sir!!
Anuj, Thank you for the positive feedback. Much appreciated.
Thank you. You are so clear whenever you are explaining!
Glad it was helpful!
Sweet...... You make it impossible to fail physics thanks a lot!
I thought my headphones were broken on one side....
AM HAPPY MY HEADPHONE IS FINE
Very helpful series that helped me hash out a couple sticking points I had. Time for my test in 2 days! Thanks.
You´re an amazing teacher.
Thanks and greetings from Argentina.
Thank you and welcome to the channel!
Isn't it generally accepted that the sign convention for torques is that counter clockwise torques are positive and clockwise torques are negative?
Yes
'lil mg' would make a great rap name!! lmao
Why would you adjust the distance from the axis of rotation? This distance from the pivot is always the distance from the pivot. It's the fact that the forces are applied at an angle that's not normal to the separation vector that necessitates the use of sine and cosine. The axis of rotation and the separation vector are orthogonal, and only force that is mutually orthogonal to these vectors will provide torque. It may be the same in the end, but it is conceptually wrong.
Ryne, not sure what you mean by: "why would you "ADJUST" the distance from the axis of rotation?" The word adjust is a very general word. What specifically would you like to know?
Thank you so much! You have helped me understand a similar problem that I have been working on!
isn't it divided by cos of 15 degree for part b
It's just a mistake
Yes. But at least he erred on the side of safety. Dividing by cos(15deg) would have made Fw even smaller.
Yeah it should be cos15 not cos30
His answer 249.5N is still correct because he just multiplied by tan(15deg) in his calculator. I really wanted the ladder to slip though!
hats off you are really great ......education must be free
I didn't understand why there was no force of the ladder on the wall. Now I see that if there was no friction on the ground there would be no force of the ladder on the wall (it would just slide out to the left), so the friction at the ground is the only force to consider going toward the right
Thank you! I was wondering the same thing and couldn't figure out why there was no 'normal force' of the ladder onto the wall.
you're a life saver
Glad you found the help you needed on this channel. 🙂
In the last equation you've written the cosine of 30 degrees instead of 15 degrees. But you fixed that in your calculator by entering the tangent of 15 degrees so it doesn't really matter anyway,
Thanks so much for what you do, Michel.
you really are an excellent professor
sir,your tutorial videos are very helpful it help me to be clearly understand with my engineering subject...
You videos are very helpful , thank you soo much
All your videos are useful thankyou we expect more
Thank you sir, you are a great teacher. I have subscribed your channel, it's really helpful.
Thanks man. You da real MVP
Ty sir and cool bow tie
YOU ARE A LEGEND MICHEL VAN BIEZEN. A FOOKIN LEGEND
why should ladder be put at 75 degrees?
Mr. Michel, one thing really bamboozles me that why is the normal force from the wall 🧱 perpendicular to the wall, while the normal force from the ground in parallel to or in the direction of the ladder and has components, which the force from the wall didn't have. Please educate me why is that so?
When there is friction the force of the surface pushing back with not be perpendicular to the surface, But if there is no friction then the force of the surface pushing back MUST be perpendicular.
@@MichelvanBiezen Thank you! I could sense it but why is it so?
THANK YOU SIR FOR CLARITY
Thank you so much!!! This video was super helpful!
You are welcome. Glad you found our videos. 🙂
In calculating the Force net in x, why was the Force of the wall no in included in it? Shouldn't it be Fnetx = (0.4)(Force normal) - (0.3)(Force wall) ?
Sir, isn’t normal force of the ground orthogonal to the force exerted by the wall?
As indicated in the free body diagram, Fnet of x = Fu - Fwall. When Fnet of x is negative the ladder begins to slide.
well learnt. very nice lecture
Glad it was helpful!
in 13:21 , why should we divide the equation with cos 30 instead of cos 15? or.. is it an error?
That is an error indeed.
You are the greatest! Thank you so much!
amazing as always. thanks for your great explaining, and Thanks alot for sharing these videos on youtube.
I dont agree with the friction force being $F=mu*N$. Friction Force is an unknown. That is the maximum force allowable before the friction force is no longer able to sustain the ladder. In other words, you are assuming that you are on a limiting condition: the ladder is hanging by its nail, so to speak. Other than that, very clear explanation.
no, if you assume there is no friction on the wall, as he states at the beginning, the reaction of the wall must be perpendicular to the wall
Gracias
Can you show an example where you work the same problem from two different pivot points to get the same answer?
The pivot point us usually chosen to eliminate one of the unknown torques or to eliminate multiple unknown torques. In this case you don't know the force between the floor and the ladder and you don't know the friction force, so picking the pivot point at that location makes the most sense.
this video has helped me a lot ...thanks a lot man ...keep up the good work :D
I'd like to give the 2 knuckleheads that disliked this video an AP Physics C test and laugh at them while they fail it. If only they would have liked it, they probably would have passed. Love the videos, I recommend them to all my struggling physics friends!
You have a mistake given in letters b sir , its supposed cos 15 , not cos 30
Yes you are correct, good catch. That means you understand it and are paying attention. (See the title of the video).
Is it correct to assume that ∑Fy = 0 (as is done at 3:33)? The question asks whether or not the ladder is slipping. If the ladder is slipping, then the ladder's center of mass is accelerating downward and ∑Fy ≠ 0.
If the ladder's center of mass is accelerating downward, then the normal force would be smaller than the calculated value of 1078 N, right? N = 1078 - (m*ay) where ay is the magnitude of the y acceleration.
The sum of the forces ONLY add up to zero, if there is no acceleration. Fnet = ma
Michel van Biezen You say (at 3:31): "For part a, we can say that the sum of the forces in the y direction must add up to zero." But how do we know that the forces in the y direction add up to zero? To ask it another way: how do we know that there is no acceleration in the y direction?
To put it a bit more pointedly: the ladder propped up against a wall can't accelerate leftward without also having its center of mass accelerate downward. (And eventually the ladder accelerates far enough downward that its center of mass hits the floor.) In your solution, you assume that ∑Fy = 0. This is commensurate with assuming that ∑Fx = 0, since the ladder can't accelerate leftward without also accelerating downward. But if we assume ∑Fy = 0, there's no need to determine whether ∑Fx could equal 0.
So, when we say that ∑Fy = 0, we assume that the ladder ISN'T slipping.
Perhaps the approach in your video is this:
(a) assume that ∑Fy = 0
(b) from (a), we conclude that ∑Fx = 0
(c) from (b), we conclude the maximum static friction is µN and µN > Fwall
(d) by calculating µN and Fwall, we can show that µN > Fwall
(e) since (d) didn't yield a contradiction with (c), the assumption from (a) must be correct
Is this the line of reasoning you're using in the video? What's the basis for assuming that the net force (in both directions) is 0, and once you assume ∑Fy = 0, why would you check to see whether ∑Fx = 0?
jdmphys
Make the assumption that the ladder doesn't slip.
Work out the problem and see if your assumption is correct
Ur amazing, thank you so much!
Glad you found our videos.
michel van biezen you are my hero i love your videos
The thing that I do not understand in all of these problems is why they calculate the force of the ladder to be using the mass of 80kg for the mass of the man on the ladder when the ladder is also a mass so therefore wouldn't the mass of where the man is standing have the combined masses of the ladder and the man?
+Mitchell Arnott
If you take the mass of the ladder into account, you must add another force vector (the weight of the ladder) acting through the center of mass of the ladder (half way up the ladder).
+Michel van Biezen sir you explain better than my teacher. but I still dont understand how to determine what forces needed for the problem.
Sir please make lectures on instantaneous axis of rotation
I have a problem, why the wall does not provide upward reaction (friction) in this case? Thank you.
In this example the friction between the ladder and the wall is zero. (frictionless wall). In many cases that is realistic as the wall tends to offer little friction if the wall is smooth. We have examples where there is friction on the wall.
Thank you for the video! It was very helpful. I was wondering why you only took the horizontal component of the distance to compute the torque. Shouldn’t you use the magnitude of the distance vector between the pivot and the center of mass? Not the horizontal component of that vector?
There are several ways in which you can do it. You can use vector products, or the direction of the force vector x cos (theta) or the method shown here.
Thank you!
You're welcome! 🙂
Suddenly got curious about this problem while climbing up the ladder to retrieve my baseball...guess I'm years late!
Audio only on left side :(
Love your videos. Totally helpful!
Great video. very clear explanations
thank you for this lecture. It helped me a lot...
thank you perfectly explained
sir isn't 1/3 supposed to be 2/3 since the pivot is at the other end
Thank you so much. You helped me a lot.
Great problem... I was close, but I didn't see that the vertical wall was frictionless. D'oh! I was treating it with the same coefficient as the floor. Bugger me.
Why wouldn’t the normal force exerted by the wall be equal to the friction force when we sum the forces in the x direction
The friction force between the ladder and the floor have an upper limit. If the normal force by the wall exceeds, the maximum friction force, the ladder will slip.
@@MichelvanBiezen thank you, physics need logic and reasoning before math which is something I lack
That is correct. Understanding physics includes knowing how to interpret the laws and definitions of physis and then learning how to apply them.
Very helpful. All agree!
I have a question. I learned from Khan Academy and such to find the torque by multiplying the distance from the pivot by the perpendicular component of the force. I suppose multiplying it by its “shadow” like you did is equivalent?
There are different ways to calculate the torque. You can find all the details here in this playlist on this channel: PHYSICS 15 TORQUE
thank you so much, it helped me a lot, but i think at the part b, it should be all over cos15 degrees :)
Why when calculating sum of Y forces we don't multiply the ladder and guy's weight by sin(theta)?
When we calcuate the torque we multiply the force (the total force) with the moment arem (= the perpendicular distance from the line of action of the force to the point of rotation).
@@MichelvanBiezen Wow, you still respond after so long! Thanks! So for torque we only care about the perpendicular force (constructing a new triangle) relative to the ladder, for force on the wall that would be sin theta and for the ladder + guys weight cos theta.
What about for the x-y components near 4:21? We don't use any thetas here, just mg?
We have a whole playlist the explain the fundamentals of torque, starting with this video: Physics - Mechanics: Ch 15 Torque Fundamentals (1 of 13) What is Torque? czcams.com/video/KyZ1-fzcng8/video.html
@@MichelvanBiezen great, ill check those out for my assessment tomorrow. Does this playlist include, tipping, seesaws, beams by any chance?
Isn't the force of the wall on the ladder canceled out by the force of the ladder on the wall, so the net horizontal force is only the friction force? And how is the net torque zero when the ladder is moving? I guess rotation and horizontal displacement aren't related, but it's kind of hazy to me.
+AquaFairy9
The force of the ladder acts ON the wall.
The force of the wall acts ON the ladder.
They cannot cancel out unless they act on the same object.
+Michel van Biezen
Okay, thank you. But I'm also wondering now, isn't the weight of the ladder and person also acting ON the ground, but they're also forces on the ladder?
More generally, I'm wondering...if all of the weight of the person and the ladder are in the downward direction, where did the horizontal force against the wall come from?
+AquaFairy9
Newton's third law.
It is the reaction force to the friction force.
Thank you very much, youre a good teacher 🌼😊
this was a great help, thank you very much:)
How do you choose the direction of the normal force? I am a bit confused because even though the situation is the same, in some cases it is perpendicular, and in some cases it is at an angle
The "normal" force is always perpendicular to the surface. (that is why they call it the normal force)
If the angle was not given, how would one go about solving it this same question ?
Not clear why Fw needs to be less than the force of friction. An explanation would be appreciated.
Since those are the only two forces acting in the x-direction, if Fw is greater than the friction force, then there will be a net force to the left and the ladder will slide.
Thank you so much
i believe your supposed to be dividing by cos 15 still??
Will,
You are correct. It should obviously be 15 degrees (not 30 degrees).
Michel van Biezen and d3 become 0 angle become 0 with respect to the climber at the top of the ladder .very good exercise..
thanks for clarifying that... i was like what!
When setting the x components of the forces why didnt we account for the force of the ladder on the wall . furthermore, why isnt Ffriction= Fwallonladder-Fladderonwall?
Note that the coefficient of friction on the wall is zero.
Hi Michael, since we have static equilibrium, shouldn’t the force of friction (Ff) equal the reaction force of the wall (Fw)?…..otherwise as Ff > Fw, then there is a net force in the horizontal direction to the right equal to
Ff - Fw and therefore the system cannot be in equilibrium….unless I am missing something here?….
Thank you 🙏
That is correct. The friction force will always equal the force causing the friction force, untill the force exceeds the maximum possible friction force. To find that limit (as done in this problem), we use the maximum possible friction force.
@@MichelvanBiezen Thanks Michael….so here the friction force, Ff, is a reaction to the normal Fw….ie Fw is causing Ff and as long as
Fw < Ffmax there will be equilibrium?
Thank you
@@MichelvanBiezen Sorry Michael….just one more point if i may please…even though the max friction force = 323N, when the person is at position L/3,
Fw = 109N….
therefore Ff = 109N ?
Thank you so much 🙏
That is correct.
@@MichelvanBiezen Thank you Michael 🙏
Question, if the vertical wall had a non-zero coefficient of static friction, would we have considered that force of friction in the net force in y-direction calculations? My thought is yes, but wanted to check to make sure that I'm on the right track. Thx.
Yes you would consider the friction force in the y-direction. There are several examples of that in this playlist: MECHANICAL ENGINEERING 11 FRICTION
Michel van Biezen thx!
Sir, please make a video on the case of the cat on a plank because the question says that the plank BEGINS TO TIP. Because I don't understand what is the meaning of the PLANK BEGINS TO TIP.
If I am understanding correctly - unless you are given an angular velocity (or any more specific information) it means that the plank is still in static equilibrium but JUST so. Meaning that if the cat were to move a hair further, or weigh an itty bit more - the plank would tip. This allows you to set it up as Tnet = 0 and solve for the unknown value - the mass of the cat, the distance of the cat from the pivot point, etc.
Thanks man, You got it right but I had sorted it out before.
5:54 Clockwise torque is positive? Is it supposed to be negative???
Oh, I got it. He then said we can do it the other way. That's the way how he does it.
When we only consider the magnitude of the torque, it doesn't matter. If we consider the vector quantity of the torque, then clockwise is indeed negative.
Man on Ladder
Ladder: I'm about to end this mans career (literally)
concept cleared....
Is there anyway to solve for the mass by only knowing the angles in which the man is pushing against the ladder? i.e. angle of the ladder against the wall, height of the wall, length of the ladder and the coefficient of friction.
PLEASE DO AN EXAMPLE WHICH THE ANGLE OF INCLINATION AND LENGTH OF LADDER ARE NOT GIVEN. BUT, FRICTION COEFFICIENTS OF BOTH WALL AND FLOOR ARE GIVEN. THERE IS WEIGHT OF THE LADDER AND THE ZOMBIE IS 3/4 WAY UP THE LADDER. HELP ME GET THE EQUATION FOR ANGLE OF INCLINATION.