Hey Micheal, I've watched a bunch of your physics videos and currently in college physics. You may or may not read this but if you do, I just want to thank you so much for making these videos because they have really helped me a lot.
This is a fantastic video. However, a fun fact that I would just like to clarify that the biceps brachii does not in fact attach anywhere on the humerus (:
amazing video and explanation. This force is if the weight stayed at a fixed angel, could you explain or tell me what I could research to figure out the force at each degree from 180 to 0 without having to do 180 different equations ?! awesome explanation of the size of the universe in a recent video btw!
Austin, You are correct. If you are only calculating the magnitude of the torque, you can define it anyway you want (like done in this video). However if you want to express it as a vector, one should define it as you indicated to be correct.
I think rightfully to calculate moments or torque, you HAVE to (cannot emphasize this enough) use Force times Perpendicular Distance. That is the rule. If there is an angle, no matter how small the difference the answer is. It is not right to ignore angles if there is. Otherwise, this is a good video
I just had this discussion with my gf, conceptually. Then I dusted off my old physics book. So it seems to me that a longer arm will require more torque to lift the same dumbbell. Is that the right concept? And is the work done equal to zero because the acting force will always be perpendicular for the dumbbell? Thank you so much for the video!
So this means someone with a longer forearm needs to generate higher torque(or force output from muscle) than someone with a shorter forearm to lift the same weight in a curling movement. Nice video.
That probably helps explain the skinny but strong guys phenomenon. They have better leverage. I suppose if the bicep attached further out on the arm that would help too.
Now this is a very useful physics video. I wish it was easier to find well explained body mechanics stuff. Is there a torque video that would help with calculating the force if you were lifting an object with your arm muscles on the other side? Like if you were bench pressing instead of curling.
Now this is a pretty useful physics video. Is there a video that goes over what you need to calculate force from arm muscles on the other side? Like if you were bench pressing instead of curling weight?
In this case we are using pounds (lbs) which is a unit of force and therefore represents weight. 1 lb= 1 slug x 32 feet/sec^2 (like 1N = 1 kg x 9.8 m/sec^2) and that is why we didn't need to multiply by g.
Hi , i heard that Counterclockwise is the positive rotation direction and clockwise is the negative direction. For example, a torque that rotates an object counterclockwise is a positive torque … so in the video it’s totally the opposite and iam kinda confused now
If considered as a vector quantity, you are absolutely correct. For torque as a vector, ounter-clockwise is indeed positive. If not considered as a vector there is no problem with reversing that, but in hindsight, I should have remained constent with the vector definition to avoid confusion.
Hey, Michel thanks for the video I have a question about this problem. Torque is equal to force times the perpendicular distance. The contraction force of the bicep is not perpendicular to d3. Nor is d3 directed to the pivot. So I'm just wondering whether it is Force*cos(theta)*d3 and we just resolve the component for the force to be perpendicular to d3 ?
The angle is so small that it can be ignored. (take the cos of 3 degrees and you will see), but you are correct that if the angle is significant you should defenitely include it.
Hey Michel, What happens when you take the small angle into consideration in the calculation, what happens? Also what if the bicep was not perpendicular to the line of gravity, so lets say the forearm and bicep make a 70 degree angle ( the top of the bicep cur)?
If I could bicep curl a 20kg and is 6 x further than the bicep point then the bicep could lift 120kg which is quite strong for a single little muscle. as strong as my max deadlift. two of them could lift 3 x my weight. I wonder what the leg muscles can do. must be 300 to 400 kg force
Its more than one ton for average gym rat cause femur is longer and stronger. World record is 500 kg plus. So that should be around 12-14 times. Bicep in this case was more than 9 times:)
@@SAMARTHNAIKDHURE_ I reckon two quads and two glutes could lift a small car direct on the attachment points (If you could take the muscles out, but still wired up and tied the ends to the car and a frame), though the range of travel of the muscle is probably less than that needed to raise the suspension enough to get the wheels off the ground. That of a strongman could probably lift an SUV. If I did a squat of 120 kg and the muscles are on a 7:1, then for me it's 7 x 120 kg + 7 x what my upper body weighs (maybe 50 kg) so 1190 kgf. For an 8:1 ratio, that increases to 1360 kgf.
I had this same type of problem on my pearson my lab homework........Why did you half the 35cm to 17.5cm for d2? I was getting it wrong on my computer homework....When I half the 35cm and did the calculation, I got it right. Why did you do that?
hey, (in aus we use metric) the formula for torque i was given was T=F * r sin(theta), but you seem too just to the force multiplied by the distance without the sin(theta), is this because in your example everything is vertical and sin90 is equal too 1
okay I can curl a 40 pound just as much as i can bench press about 150 pounds / 2 = 75lbs per chest muscle. Bench pressing is a linear rep, no rotation hence no torque. So does this technically mean that my biceps, although a smaller weight can produce more force then my individual chest muscles? And if so, can it be possible that there is still more energy required to bench vs the curl?
Bench pressing is not done with the chest muscles (although they are involved). It depends on the strength of the muscle doing the contracting and the moment arm of the torque required for the muscle.
@@MichelvanBiezen you think it’s linear but it isn’t, your upper arm is rotating around your shoulder joint while your forearm rotates around your elbow, both act together to make it appear linear
Plz I have a problem in the finding of distances in the torque and how can we know that we should use the law of rotation(torque) or newton's laws 1&2?!
So by inverting that, we can estimate that 35/4 * ? = ~350 So to calculate force exerted by a mechanical finger, with 1st finger segment from the palm being 2" long, with force being exerted to pull that finger at 1" of its length, then dividing the total exerted force ON that finger by the resultant number should give a rough approximation of the force exerted at the end of that finger. In my problem, a 2" mechanical finger segment has 2,700lbs of force exerted at an attachment location of 1" down that finger segment, so directly at the midpoint. 2 divided by 1 = 2, so 2,700lbs divided by 2 = 1,350lbs total force (roughly) exerted at the end of the finger segment.
Then you have to multiply the tension of the biceps with the cosine of the angle. For example the cos (4 degrees) = 0.99756 , which really doesn't make any significant difference.
Black Widow You are correct if you want to express the answer as a vector. If you only want the magnitude of the answer, like in this example, it doesn't matter. I picked the sign that would give me a positive answer. But strictly speaking when dealing with vectors, you are correct.
Michel van Biezen ah. Alright didnt realize that , very sorry im taking ohysics for the first time and torque was a new concept for me so was a bit confused is all .Thank you
Yes. If there is no motion (or more precisely no angular acceleration) then there is no net torque and all the torques added together must add up to zero.
not sure if i made a mistake or not, but it seems as though you did not multiply the wieght by gravity, is that becuase you used pounds and cm, and gravity cancelled out?
+Tyler Murphy That is the exact situation in this example problem. The forces are the same for a static situation as they are under constant velocity (if no friction is involved)
It's so mad how my teacher ignored so many informations you wrote , I think it's was going to be so much easier with those informations, but teacher was like you've already learnt this in high school lol x
Ashkan Class Look at the other examples. The method is the same. Remember the sentence. "the perpendicular distance from the point of rotation to the line of action of the force"
@@MichelvanBiezen maybe if they swung it using their legs =) My arms were pretty nice when I was doing five sets of forty. They weren't huge by any stretch of the imagination. If you grappled with me, you would probably notice them.
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Hey Micheal, I've watched a bunch of your physics videos and currently in college physics. You may or may not read this but if you do, I just want to thank you so much for making these videos because they have really helped me a lot.
+Nathan “Wild Rain” Nguyen
You are quite welcome. I am glad I can give back a little to the world.
Keep up your studies, it is the key to the future.
@@MichelvanBiezen you are giving a lot think you very much
how is your life now?
I thought he was good but then in he brought in medical related problems and now i think he is great. Relates so much to my degree.
Ahhh! Please newtons and kgs!!
+Luke Saliba We are working on some additional torque problems that will more of the metric units in them. They will be up in a few weeks.
I just found my next favorite Physics teacher.
This helped so much, I never understood the bicep problems..
Thank you!
Yes, I am always amazed how the human body works in a mechanical sense.
I am proud of my biceps, they are stronger than I thought 😀
Great lesson 👏👌
This is a fantastic video. However, a fun fact that I would just like to clarify that the biceps brachii does not in fact attach anywhere on the humerus (:
I've been thinking about this for a few days now. Glad I was able to find this video.
Glad you found us
I honestly thounght the strain would be higher. Then again, everyone's arms are different.
all u r lectures r par excellence - i have learnt a lot
Thank you so much for all your videos. They are much appreciated!
amazing video and explanation. This force is if the weight stayed at a fixed angel, could you explain or tell me what I could research to figure out the force at each degree from 180 to 0 without having to do 180 different equations ?!
awesome explanation of the size of the universe in a recent video btw!
Set up the same equation, except that the angle will now be unknown.
I was looking to put more science into my workouts and this was just simply an amazing video that helped me a lot! Thanks Michel!!!
Bro same using it for research in sprinting :)
how do i calculate this if the angle between the forearm and biceps is not 90
i was preparing my test in university thise videos also help me during study time
Dr.michlel why the angles are not stated as the equation says f*d cos( theta) in this situation??
And thank you
You can use either method to calculate the torque.
I thought clockwise is negative and counter clockwise is positive...?
+Micheal van Biezen
Austin, You are correct. If you are only calculating the magnitude of the torque, you can define it anyway you want (like done in this video). However if you want to express it as a vector, one should define it as you indicated to be correct.
Can you also show an example of question applying torque concept, except not lifting a weight but instead pushing a wheelchair by the hand rim?
We have a number of examples for torque with the body. Did you see the other ones in the playlist?
Llevo 2 minutos de video, pero este video es increíble, muchas gracias, saludos desde Colombia.
Thank you and welcome to the channel!
I think rightfully to calculate moments or torque, you HAVE to (cannot emphasize this enough) use Force times Perpendicular Distance. That is the rule. If there is an angle, no matter how small the difference the answer is. It is not right to ignore angles if there is. Otherwise, this is a good video
do you have any courses specifically cater for exercise movement physics?
I just had this discussion with my gf, conceptually. Then I dusted off my old physics book.
So it seems to me that a longer arm will require more torque to lift the same dumbbell. Is that the right concept?
And is the work done equal to zero because the acting force will always be perpendicular for the dumbbell?
Thank you so much for the video!
So this means someone with a longer forearm needs to generate higher torque(or force output from muscle) than someone with a shorter forearm to lift the same weight in a curling movement. Nice video.
That is correct. 🙂
That probably helps explain the skinny but strong guys phenomenon. They have better leverage. I suppose if the bicep attached further out on the arm that would help too.
Now this is a very useful physics video. I wish it was easier to find well explained body mechanics stuff. Is there a torque video that would help with calculating the force if you were lifting an object with your arm muscles on the other side? Like if you were bench pressing instead of curling.
i love your methods!
Now this is a pretty useful physics video. Is there a video that goes over what you need to calculate force from arm muscles on the other side? Like if you were bench pressing instead of curling weight?
No, we don't have that one. There are a few others, such as the force on the Achilles tendon.
Thank you so much. I love all your lectures. It is very clearly explained.
Thanks a lot sir! Please do more biomechanics videos of working out.
Thank you a lot for such helpful videos, but shouldn't multiply by g?
In this case we are using pounds (lbs) which is a unit of force and therefore represents weight. 1 lb= 1 slug x 32 feet/sec^2 (like 1N = 1 kg x 9.8 m/sec^2) and that is why we didn't need to multiply by g.
Hi please how do you calculate the force if there is an angle
Then you multiply by cos (angle) with reference to the vertical. We have a number of examples of that in the playlist.
You're assuming the barbell is a point mass too, correct?
The barbell is gripped at the center of mass and therefore acts as a point object.
I have a question about the distance 3. Should it be perpendicular to Tension line action?
The angle is so small that it is essentially zero and thus d is essentially perpendicular to the tension.
Are there any videos on levers?
Yes, look in the physics playlists on torque i
Hi , i heard that Counterclockwise is the positive rotation direction and clockwise is the negative direction. For example, a torque that rotates an object counterclockwise is a positive torque … so in the video it’s totally the opposite and iam kinda confused now
If considered as a vector quantity, you are absolutely correct. For torque as a vector, ounter-clockwise is indeed positive. If not considered as a vector there is no problem with reversing that, but in hindsight, I should have remained constent with the vector definition to avoid confusion.
@@MichelvanBiezen thank you alot
Thank you for your video! Could you please clarify as I am quite confused:
Mg = F (which gives Newtons), why do we use lbs instead?
The same equations can be used for the imperial system of units. In the US we still use pounds instead of Newtons. F = ma [lbs] = [slug] [ft/sec^2]
@@MichelvanBiezen wait do you mean america? Cause we use N and Kg
Hey, Michel thanks for the video I have a question about this problem. Torque is equal to force times the perpendicular distance. The contraction force of the bicep is not perpendicular to d3. Nor is d3 directed to the pivot. So I'm just wondering whether it is Force*cos(theta)*d3 and we just resolve the component for the force to be perpendicular to d3 ?
The angle is so small that it can be ignored. (take the cos of 3 degrees and you will see), but you are correct that if the angle is significant you should defenitely include it.
@@MichelvanBiezen Thanks a lot
Hey Michel, What happens when you take the small angle into consideration in the calculation, what happens? Also what if the bicep was not perpendicular to the line of gravity, so lets say the forearm and bicep make a 70 degree angle ( the top of the bicep cur)?
Small angles make very little difference. Take the cosine of 2 degrees and see what you get.
how do you find the force that will cause equllibruim top the system
Great video, thanks! Very helpful!
Great video!
Glad you liked it
from when did lbs become unit of Force?
In the US, the units of force are pounds. 1 pound is the force that gives a mass of 1 slug the acceleration of 1 foot per second.
If I could bicep curl a 20kg and is 6 x further than the bicep point then the bicep could lift 120kg which is quite strong for a single little muscle. as strong as my max deadlift. two of them could lift 3 x my weight. I wonder what the leg muscles can do. must be 300 to 400 kg force
Its more than one ton for average gym rat cause femur is longer and stronger. World record is 500 kg plus. So that should be around 12-14 times. Bicep in this case was more than 9 times:)
For 20 kg in this case, it would be 180 kg for bicep :)
@@SAMARTHNAIKDHURE_ I reckon two quads and two glutes could lift a small car direct on the attachment points (If you could take the muscles out, but still wired up and tied the ends to the car and a frame), though the range of travel of the muscle is probably less than that needed to raise the suspension enough to get the wheels off the ground. That of a strongman could probably lift an SUV. If I did a squat of 120 kg and the muscles are on a 7:1, then for me it's 7 x 120 kg + 7 x what my upper body weighs (maybe 50 kg) so 1190 kgf. For an 8:1 ratio, that increases to 1360 kgf.
WHY IS d3 4cm...ACCORDING TO ME d3 IS PERPENDICULAR TO THE BICEPTS
d3 is a given quantity. It is the distance from the joint (point of rotation) to the point where the bicep is attached.
so shorts arms is an advantage to curling weight?
It certainly is.
@@MichelvanBiezen or a biceps tendon that attaches further from the elbow
thanks for the video! but what about the reaction force ?
How would you account for the angle at the bicep?
What do you mean by: "How would you account" ?
I had this same type of problem on my pearson my lab homework........Why did you half the 35cm to 17.5cm for d2? I was getting it wrong on my computer homework....When I half the 35cm and did the calculation, I got it right. Why did you do that?
For the torque caused by the weight of the forearm. we assumed that the center of mass is halfway from the hand to the elbow.
What if the forearm isnt straight but at an angle of theta below the horizontal. what would the equation look like?
Then you have to multiply times the cos (angle) where the angle is between the fore arm and the horizontal.
what about g .why isn't it calculated ?
It is. Pounds is a unit of force that already includes g (line Newtons)
Michel van Biezen thank you
please id like a question why did you assume that the total torque acting on the arm equal zero
If the net torque is not zero, the arm would be moving. Thus if the arm is not moving, the net torque about the joint of the elbow must be zero.
Michel van Biezen thank you too much
hey, (in aus we use metric) the formula for torque i was given was T=F * r sin(theta), but you seem too just to the force multiplied by the distance without the sin(theta), is this because in your example everything is vertical and sin90 is equal too 1
okay I can curl a 40 pound just as much as i can bench press about 150 pounds / 2 = 75lbs per chest muscle. Bench pressing is a linear rep, no rotation hence no torque. So does this technically mean that my biceps, although a smaller weight can produce more force then my individual chest muscles? And if so, can it be possible that there is still more energy required to bench vs the curl?
Bench pressing is not done with the chest muscles (although they are involved). It depends on the strength of the muscle doing the contracting and the moment arm of the torque required for the muscle.
@@MichelvanBiezen you think it’s linear but it isn’t, your upper arm is rotating around your shoulder joint while your forearm rotates around your elbow, both act together to make it appear linear
and the distance top which it will applied
Could you do a video of torque acting on human leg
Why didn't you include the g (9.8) when you was calculating?
pounds is a unit of force and already includes (g).
What if I want to calculate this with the angle of bicep? How will I do that?
Then you have to multiply the force of the bicep by the cos of the angle between the perpendicular and the direction the bicep pulls.
Would really like to find out forces at finger (top segment, directly after palm) vs forces generated at forearm muscle.
Forearms are not your muscle group eh? 40 lbs is too much for you?
What if there is a given sin alpha=0.98 (angle alpha=105°) between the bicep and the arm which is 35 cm in ur example?!!!
why hv u takrn 4 & 35 cm pricesly .. like any reason??
The numbers we pick are not as important as the technique to solve this problem.
Thank you so much
What about the characteristics of normal reaction applied (on where u put number 1)?!
Plz I have a problem in the finding of distances in the torque and how can we know that we should use the law of rotation(torque) or newton's laws 1&2?!
Whenever something rotates about a central point (like the elbow joint) we use torque and rotational motion
@@MichelvanBiezen okk thnx
35/4*40=350~363 Very simple!
So by inverting that, we can estimate that 35/4 * ? = ~350
So to calculate force exerted by a mechanical finger, with 1st finger segment from the palm being 2" long, with force being exerted to pull that finger at 1" of its length, then dividing the total exerted force ON that finger by the resultant number should give a rough approximation of the force exerted at the end of that finger. In my problem, a 2" mechanical finger segment has 2,700lbs of force exerted at an attachment location of 1" down that finger segment, so directly at the midpoint. 2 divided by 1 = 2, so 2,700lbs divided by 2 = 1,350lbs total force (roughly) exerted at the end of the finger segment.
@@davidsirmons i didn't understand anything :D My native language is not english :D U are using so many terms.
what about the friction in the body articulations?
What do you do when you can’t ignore the angle of the biceps?
Then you have to multiply the tension of the biceps with the cosine of the angle. For example the cos (4 degrees) = 0.99756 , which really doesn't make any significant difference.
sir, is there any force on the elbow joint??
Yes there is.
Thank you
You're welcome
this was VERY INFORMATIVE thanks a lot
BICEPS !!!
khuth
lbs is not a unit for force
Yes it is.
@@MichelvanBiezen ohh right, the US measurement system hahah. Great video though, thanks!
if it asked for the answer in Newtons, would the correct answer be 1617 N?
Brian,
Since 1 lb = 4.448 N 363 lbs = 1615 N
What kind of class lever is it
Hey there! Aren't you supposed to multiply the g=9.8 as well?
franxclusive,
The unit lbs (pounds) is a unit of weight (force) which already includes g (32 ft/sec^2)
Ok! Got it. Thank you!! :)
Counter clockwise direction is positive , so clockwise would be negative no?
Black Widow
You are correct if you want to express the answer as a vector.
If you only want the magnitude of the answer, like in this example, it doesn't matter.
I picked the sign that would give me a positive answer.
But strictly speaking when dealing with vectors, you are correct.
Michel van Biezen ah. Alright didnt realize that , very sorry im taking ohysics for the first time and torque was a new concept for me so was a bit confused is all .Thank you
Black Widow
No need to apologize, that was a very good question. It takes a while to grasp the concept and meaning of vectors and scalars.
Prof, I have a question. why is Summation of Torque= 0? What is the reason for that? Is it due to the arm being stationary?
Yes. If there is no motion (or more precisely no angular acceleration) then there is no net torque and all the torques added together must add up to zero.
Alright, thank you!
hello prof, little clarifications sir, thought we should be left with a cm after dividing one by the other .
(lbs * cm) / cm = lbs (the centimeters cancel out )
thank you teacher
You are welcome
not sure if i made a mistake or not, but it seems as though you did not multiply the wieght by gravity, is that becuase you used pounds and cm, and gravity cancelled out?
+Doc Pounds is a unit of weight (like Newtons) and thus we don't have to multiply by g.
+Michel van Biezen ah, i understand, thank you for the quick reply.
Why not just say that the sum of the forces = 0? Is there some additional force that doesn’t cause a torque?
That is correct. But some problems can only be solved by using the sum of the torques. (or using the torques makes it easier)
very helpful! thankx
hallo mr. van biesen, im askink for your mail, can i get it please
We try to keep our correspondence to these comments.
what if you need to keep the forearm static?
+Tyler Murphy That is the exact situation in this example problem. The forces are the same for a static situation as they are under constant velocity (if no friction is involved)
Biokinetics of ragdolls look useful and fun. 😉
thanks
Thank you so so so much!
This is the exact question I needed help with, thank you!!!
Glad it was helpful! 🙂
It's so mad how my teacher ignored so many informations you wrote , I think it's was going to be so much easier with those informations, but teacher was like you've already learnt this in high school lol x
Thanks a lot! What if we have angle?
Ashkan Class
Look at the other examples. The method is the same.
Remember the sentence. "the perpendicular distance from the point of rotation to the line of action of the force"
thank u for the video +1 thumps up
So when I would use 80lb dumbbells at one point for 1-2 reps, 726 lbs was being generated by one arm. Wow. I had no idea...
Wait, you forgot gravity in your calculations?
Gravity was included.
i thought clockwise was negative
If used as a vector quantity yes. If used as a magnitude it doesn't matter.
So where are we in standard deviations in terms of who in this comment section can lift that 40 lbs?
I would think that most people in good shape can do that.
@@MichelvanBiezen maybe if they swung it using their legs =) My arms were pretty nice when I was doing five sets of forty. They weren't huge by any stretch of the imagination. If you grappled with me, you would probably notice them.
Bicept?
yes
BicepTs...
دكتور ممكن سؤال ع ضهر
What is the question?
Is it just me or did he spell 'biceps' wrong? twice on the board?
I certainly did spell "biceps" wrong. (English is not my native language, and some of the spelling takes a little to get used to)
@@MichelvanBiezen Sorry sir. Great video though. The excellent physics explanations is what I'm really here for :)
No offense taken. (I do struggle with English spelling, oh well, that is why I became a physicist)
wtf is a bicept
Drop the t --> bicep 😂😂
4cm × Fbiceps= 3lbs + X (?) + 35cm × 40lbs hehe
BIcEpTS
Yes the "t" does not belong there: biceps
snap snap