Thank you for posting these videos. My professor likes to show off in class and show how fast he can solve a problem and doesn't explain anything. You take your time and don't skip steps and are fun to listen to. Without these videos I'd barely be passing, and would certainly not have an A. I really appreciate the work you put in to making each video interesting and concise.
If you assume the bear does reach the basket, it's easier to just calculate what the tension is at the point and see that its is larger than the Tmax using contradiction. I did this first because i always pause your videos and try to work them out myself. Thank you so much, you're the man.
Michel van Biezen Hi Professor, I really enjoy all your videos and seen quite a bit. Is there a solid mathematical proof of why Torque is equivalent to the Force multiplied by Lever Arm? It’s really good to know that what you’re doing is mathematically justified. Please get back to my comment. Love your videos, cheers.
I appreciate you so much sir. To me I didnt need to use distance x, since I want to know if starving bear reached the basket, I just assume he is at distance L 1 and if TL sintheta is bigger than the sum of the clockwise forces than he will reach the basket.
Yes, you can do that in order to determine if he will reach the basket. But if you want to know how far the bear can go before the string breaks you'll need to use "x".
Instead of using the perpendicular distance between the tension and the pivot point, you could use the lenght of the bar and the y-component of the tension (since the x-component does not produce torque), right?
How the can we say that the x-component does not produce torque while the force of T on the plank is not perpendicular to the plank? That means torque divided in the two components x and y.
Hi Sir, I am confused about the clockwise and anticlockwise direction. Because there are videos you mentioned clockwise direction is in negative sign while some videos you mentioned is positive sign. So actually clockwise direction is positive or negative sign?
If we use torque as a VECTOR quantity, then the convention is that counter-clockwise is positive and clockwise is negative. If we use torque as a magnitude it doesn't matter as long as we stay consistent.
The way I did it is that I assumed that the bear was already on the edge and I calculated the amount of force it exerts when it's there to see if it exceeds the maximum force that the rope can withstand and I've got an amount that is larger than the maximum force allowed so I concluded he won't reach it because the rope will already snap. Is it still fine doing it this way?
but the bear "W" is still in the direction of the T (Add to T) and to the left of the cg of the plank so how it can exert a force to break the rope. Please explain a little bit more. Thanks a lot.
@@MichelvanBiezen so you mean at 16 feet the torque due to the weight of the bear (plus the torque due to the weight of the beam) would be greater than the torque due to the rope attached to the beam and therefore the system will collapse, right?
Just to make sure I'm picking this up, can someone confirm the force on T to be 236.7lbs if the bear were to be set on the end of the lever (20ft)? Thank you so much for these videos!
HelloYellow69, A normal force is a reactionary force of a surface pushing back against the original force that caused it (see Newton's third law). Thus the normal force is the force of the beam pushing back on the bear. Therefore that force does not affect the beam, nor the torque on the beam.
Amgad Hammouda In a static situation, (nothing is moving), the sum of the forces are zero and the sum of the torques are zero. If the net torque is not zero, there will be an angular acceleration
Hmmmmmmmm if the BEAR waits a few weeks, without eating, he'll lose weight and therefore can give it another shot!!.. I should figure out what the bear's Maximum Weight should be in order to Walk all the way over to the , by now, STALE picnic basket.. lol oh well... (Okay.. did the calculation and, assuming no errors, THE BEAR SHOULD WEIGH NO MORE THAN 128.2 Pounds..) ... but beware the Creaky board!!
+steven neufeld You are correct. The standard is that counterclockwise is positive. However, if all you want to do is find the magnitude of the torque, it doesn't matter.
+Michel van Biezen Alright thanks for the information. Getting in some last minute lectures before my exam tomorrow morning. Your videos are very helpful.
Julio garcia-castro Julio, Thanks, now I understand the question. It turns out that pounds are a unit of force (like Newtons) 1 lb = the force that gives 1 slug the acceleration of 1 ft /sec^2 like 1 N = the force that gives 1 kg the acceleration of 1 m/sec^2
![ILLSLICK - KILLSHOT REMIX [FULL VERSION]](http://i.ytimg.com/vi/RIK8RrqIAzE/mqdefault.jpg)
Thank you for posting these videos. My professor likes to show off in class and show how fast he can solve a problem and doesn't explain anything. You take your time and don't skip steps and are fun to listen to. Without these videos I'd barely be passing, and would certainly not have an A. I really appreciate the work you put in to making each video interesting and concise.
If you assume the bear does reach the basket, it's easier to just calculate what the tension is at the point and see that its is larger than the Tmax using contradiction. I did this first because i always pause your videos and try to work them out myself. Thank you so much, you're the man.
+Tunde O
That is a good technique.
Michel van Biezen, you are a blessing!
Hannah,
Problems can be solved in any units, as long as one stays consistent throughout the problem like you indicated.
Michel van Biezen Hi Professor, I really enjoy all your videos and seen quite a bit. Is there a solid mathematical proof of why Torque is equivalent to the Force multiplied by Lever Arm? It’s really good to know that what you’re doing is mathematically justified. Please get back to my comment. Love your videos, cheers.
he is genius+generous
best teacher
I am far from a genius, but glad that the videos are helping.
sir there is not a single day that i dont watch your videos...
the conceptual clarity is amazing and the accuracy is unmatchable
oh my goodness, I kept forgetting to account for the moment arms and only accounted for the forces! Thank you!!
Yes, that is the key to this type of problem. 🙂
The bear just got out of the shower. Silly bear. Thanks for the direct approach.
Can't believe these videos are free. Oops, shouldn't jinx it
We plan on keeping it that way, so that everyone has access.
@@MichelvanBiezen Thank you so much!!!!
I finished with physics 1 came across this video again today and now I find physics's actually interesting.
why didn't you have to convert distance to SI units? Is it because it is constant throughout the problem?
I used another approach, which is I imagined that the bear was already at the tip then I calculated for Tension force to see if it exceeds (Tmax)
That is a good way to do it, unless you want to know how far the bear can go until it breaks.
That tie is awesome! Great video!
I appreciate you so much sir. To me I didnt need to use distance x, since I want to know if starving bear reached the basket, I just assume he is at distance L 1 and if TL sintheta is bigger than the sum of the clockwise forces than he will reach the basket.
Yes, you can do that in order to determine if he will reach the basket. But if you want to know how far the bear can go before the string breaks you'll need to use "x".
Great job sir you're videos are really helpful and the examples you give are very unique that opens up new ways of understanding the topic thank you
A pivot point is also known as a pin.
When I was working this problem I was hoping the bear would be able to get himself the picnic. I feel bad for him lol
Yes, we feel sorry for the bear.
I don't think the bear can get skinnier. lol. Thanks for the lesson. :D
So x is the max distance the bear can pass before the beam collapse? Thanks a lot sir
ha haa Starving bear............ Professor your lectures are amazing, thank you.
Thank you Professor!!
Instead of using the perpendicular distance between the tension and the pivot point, you could use the lenght of the bar and the y-component of the tension (since the x-component does not produce torque), right?
ProjectD100
you are correct
How the can we say that the x-component does not produce torque while the force of T on the plank is not perpendicular to the plank?
That means torque divided in the two components x and y.
thank you so much for these videos!
Why use d3 instead of L for the distance from the axis to the tension?
You need to use the perpendicular distance from the line of action to the point of rotation (or pivot point) when you use this technique.
Hi Sir, I am confused about the clockwise and anticlockwise direction. Because there are videos you mentioned clockwise direction is in negative sign while some videos you mentioned is positive sign. So actually clockwise direction is positive or negative sign?
If we use torque as a VECTOR quantity, then the convention is that counter-clockwise is positive and clockwise is negative. If we use torque as a magnitude it doesn't matter as long as we stay consistent.
The way I did it is that I assumed that the bear was already on the edge and I calculated the amount of force it exerts when it's there to see if it exceeds the maximum force that the rope can withstand and I've got an amount that is larger than the maximum force allowed so I concluded he won't reach it because the rope will already snap. Is it still fine doing it this way?
It is a good method. The only information that you will not get from that method is how close the bear can get before it breaks.
great vid. made me smile, ok hoot with laughter. sometimes, physics is so serious. nice to smile while learning. :)
cant you just use
(Tension*sinθ(the component for force that is perpendicular to the beam*(Length of the beam),
for the torque?
there are often multiple ways to solve a problem. Try it and see if you get the same answer.
why wouldn't you plug in 9.8 for g in the equation?!?
That is because the "weight" was given not the "mass". w = m*g
Thank you so much for replying !! I've been going through all your videos !! such a big help. Thank you.
wouldn't the bear have a normal force
Why does x=16ft means that the bear won't be able to reach the basket??
+FST NUKEZZ If the bear goes any farther than 16 feet, the torque caused by the bear will be so large that the rope will break.
Because there's only 1 counter clockwise torque and it doesn't exert enough force to hold the plank still
but the bear "W" is still in the direction of the T (Add to T) and to the left of the cg of the plank so how it can exert a force to break the rope. Please explain a little bit more. Thanks a lot.
you’re amazing 😭🙏🏿
Why didn't you multiply your mass by gravitiy when solving it?
Walter Chavez
Pounds is a unit of force, not mass.
but how do you know what value for length will the system collapse?
Because the torque will produce a force on the rope greater than the strength of the rope
@@MichelvanBiezen so you mean at 16 feet the torque due to the weight of the bear (plus the torque due to the weight of the beam) would be greater than the torque due to the rope attached to the beam and therefore the system will collapse, right?
Sir, why you didn't multiple the mass by gravity force
pounds is a unit of force (it already includes g)
Just to make sure I'm picking this up, can someone confirm the force on T to be 236.7lbs if the bear were to be set on the end of the lever (20ft)?
Thank you so much for these videos!
You are correct. If the bear was standing at the end of the beam, the tension in the rope would be 236.7 lbs.
The bear should walk to the end so the cable snaps while holding on, then the food will fall to the ground, and he can eat it on the ground.
if i assume bear will get basket at max of length and find T
so i get T>T(max)
can i do that too?
Dan niee
Definitely. That is a good approach as well.
why do you not account for the normal force of the bear on the beam in a problem like this one?
HelloYellow69,
A normal force is a reactionary force of a surface pushing back against the original force that caused it (see Newton's third law).
Thus the normal force is the force of the beam pushing back on the bear.
Therefore that force does not affect the beam, nor the torque on the beam.
So, all the previous examples the sum of the torques are zero, when exactly aren't they zero? Thank you
Amgad Hammouda
In a static situation, (nothing is moving), the sum of the forces are zero and the sum of the torques are zero.
If the net torque is not zero, there will be an angular acceleration
Michel van Biezen Thank you very much for your reply. Now I get it.
is there example that the angular acceleration(not = 0)
Feel sorry for the bear! 😅😅😅
The bears should be not climbing poles and stealing picnic baskets anyway. Ranger Smith would not like it.
😅😅😅😅😅😅
Did you cancel gravitiy along with lbs?
I'm pretty sure lbs is weight and weight is mass X gravity
You are very good but I think that you are really fast so can you be more slowly in the next lectures
Thank you. It is very helpful and funny :) You are the best :)
Those stick figures always strike me as absurdly cute-looking.
Laurelindo,
My artistic talent never made it past 1st grade. (Art was my worst subject in school).
Michel van Biezen I actually find the stick figures pretty charming, so they are a pleasant addition to the diagrams. =)
THANK YOU
nice.. very nice..
Thank you.
Hmmmmmmmm if the BEAR waits a few weeks, without eating, he'll lose weight and therefore can give it another shot!!.. I should figure out what the bear's Maximum Weight should be in order to Walk all the way over to the , by now, STALE picnic basket.. lol oh well... (Okay.. did the calculation and, assuming no errors, THE BEAR SHOULD WEIGH NO MORE THAN 128.2 Pounds..) ... but beware the Creaky board!!
+Philip Y ...And the bear has to be smarter than the AAAAverage Bear!
why does he not use the right hand rule in any of his torque videos? counter clockwise should be positive and clockwise negative.
+steven neufeld
You are correct. The standard is that counterclockwise is positive. However, if all you want to do is find the magnitude of the torque, it doesn't matter.
+Michel van Biezen Alright thanks for the information. Getting in some last minute lectures before my exam tomorrow morning. Your videos are very helpful.
this is the best lecture i have heard
Amarjeet, Thank you for the feedback
5.23-7.48
Sin
🤣
5:23
can i take d3= (L*cos theta ) ?
Jana,
No, since d3 is opposite to the angle.
By definition, the side opposite to the angle = hypotenuse * sin(theta).
am from zambia...weight conversion to newton
1 lb = 4.448 N and welcome to the channel!
What about gravity?
Julio,
Not sure what you are asking. Can you expand your question some more?
Sorry, at the 8:06 mark, for some reason thought MG was not transferred down in the equation but figured out the lbs is M*G.
Julio garcia-castro
Julio,
Thanks, now I understand the question.
It turns out that pounds are a unit of force (like Newtons)
1 lb = the force that gives 1 slug the acceleration of 1 ft /sec^2
like
1 N = the force that gives 1 kg the acceleration of 1 m/sec^2
Legend
2:26
teddy bear
Yay imperial units :D
you're cute