Physics 15 Torque Example 4 (4 of 7) The Diving Board
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- čas přidán 28. 08. 2024
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In this fourth of the seven part series I will show you how to find the forces on the diving board with a diver at the end of the diving board.
This is hands down the best explanation for the torque problems.
seriously tho
I love this guy. He explains static and equilibrium mechanism very clear and simple way with showing examples. Thank you.
I have completed my physics1 and 2 already, but I still love it because of you. Thank you so much. You deserve more subscribes. I was always the top 1 score in my class for every exam in both Physics 1 & 2 just because of your videos. Thank you!!!!
I found this lecture more interesting than my own professor at class. Life savers.
sir please help me out in this problem:- why didn't you consider the normal forces of the supports,diver and that of the board while calculating the net torque and does F2 account for all the reaction forces in the y direction and I think that F1 must be the weight of the support acting downwards
its because the distance of normal force from the point of rotation is zero
and the formula for torque is F multiplied by the distance between the line of action force and the point of rotation
thank you from gettysburg pa
You are welcome. Gettysburg is a famous place in the US
Thank you so much, this made so much sense!
You're so welcome!
thank you!!! this helped me so much for my physics exam, you explained the problem so well :)
Very well-detailed explanation of the physics of springboard diving!
Why didn't you write out any normal forces that's acting on the object? does normal force do torque?
I don't understand how you got the half meter. But I love your videos they are the reason why I'm passing physics! :)
delhely94,
When you solve these types of problems you can choose the pivot point anywhere you like. In this case, (since we are looking for F1), I picked the pivot point at point 2. Then you have to find the perpendicular distance from the line of action of the force to the pivot point. For the 60kg diving board, the distance from the center of mass of the diving board to the pivot point 2, is 0.5 m.
Michel van Biezen thank you! :)
Nobodies going to mention how he drew a perfectly straight line freehanded? #respect
If you could please explain why you did not account for the normal force acting in the positive y direction for both the fulcrum and the person on the end, that would be greatly appreciated
Only consider external forces acting on the board.
good explanation for torque forces
best explanation very clear and easy to understand , thanks professor
hi how would you shift the pivot point to the most left position please to calculate the force acting by the fulcrum?
my left ear really enjoyed that
you do excellent work sir thanks so much.
PHY2AW i think counter clockwise is positive and clockwise is negative??? Can you explain? Thank you
When calculating vectors, the direction is indeed positive in the counter clockwise direction. When calculating the magnitudes only we can choose any direction and we'll get the correct answer.
where can i donate money to support you sick bowtie swag?
I love his bowties they’re cute 🤗🥰
Thank you for this video
You are welcome. Glad you liked it. 🙂
The 1 m long diving board 1 with a diver of mass 75 kg standing at its tip oscillates
with a frequency of 3 Hz. When the diver is standing still, the amplitude of oscillation is observed
to decrease from 150 mm to 80 mm in 10 cycles. The mass of the diving board can be neglected.
1. What is the modulus of rigidity EI for the board and what is the damping ratio?
2. To which value will the initial amplitude of 150 mm decrease in 10 cycles, if the diver weighs
only 50 kg?
Note: The shear stiffness k of a cantilever beam is obtained as k =
3EI/L3
If its cantilever then ,How can we solve it?
thank you .its perfectly clear
Beautifully explained!! thanks a mill!
Glad it was helpful!
I just took my physics final and had a question similar to this. However, it said the mass of the board was negligible. I had the correct setup but all that was provided was the lengths of the board and mass of the person (44kg). Was I supposed to assume (based on physics knowledge) that the mass of the board was 2*44kg? Since this will be equal to the normal force. It was also a fulcrum involved that wasn't in the center. I hope I explained this well enough for you to assist me.
Lillian Copeland I know it’s late but for future reference, if it tells you that the mass of the board is negligible, it means that the mass of the board does not exert a force (and therefore torque) on the pivot which it normally would due to gravity. Therefore you just ignore the mass of the board and use it to get the distance from the person to the pivot point.
I am curious how it works if you took the end of the board as the pivot point how it is calculated?
Try it and you'll get the exact same answer. You can place the pivot point anywhere, but it is recommended that you place it so that one unknown can be eliminated.
Awesome explanations!
Very clear explanation. I was using this to help me determine the lbs/in of a compression spring I need to replace an actual diving board. With my measurements being: F1=55kgx9.8x1.46m+20kgx9.8x.4m÷.67m ...F1=1291.55N F2=2026.55N . Converted 2026.55N to 1495 lbs.ft.?? I figured the compression rate of a compression spring to be 17940 lbs.in. I know this is incorrect, but I don't know what I am doing wrong. Can you help plz?
since f1 and f2 is a ccw, when solving for f2 shouldn't it be 0= -f1+mg+mg-f2 =
f2=-f1+mg+mg which results in f2 being 245 ??? am i wrong?
+Prisca Love
The answers in the video are correct.
Why does the pivot point act in a positive direction?
Think about it this way: What would happen if you removed the pivot point?
This video is great. Thank you!
There were no normal forces acting on the supports?
Yes, but if we take the pivot point to be at that location, they do not add any torque to the equation.
Michel van Biezen Ah, I see. Thank you :)
why isn't F1 pointing up? I thought F1 was supporting the weight of the diver
Try it with a ruler and an eraser on one end. If you don't push down on the ruler on the other end, the eraser will fall down.
thank u for this 💕
+Michel van Biezen
How did you pick the point (1/2) for d2, when solving at pivot point 1.
The distance from the end to the pivot point is 3, but I don't understand how you got d2, if you didn't know the length of the board
+Karen Roman
We assumed that the board is uniform, so the center of mass is at the middle of the board.
@@MichelvanBiezen then doesn't that mean that the perpendicular distance for the torque acting on the board would be 2.5m instead of 0.5 when the pivot point is at pivot point 1?
How is it that the fulcrum applies a Normal force, while the other locations (eg. suspensor on the far left of the diving board) does not?
On the far left, there is a force acting downward to keep the board from swinging upward.
okay sir, so how to find the normal force at point A?
Find yourself
Sum up all the forces in the y direction.
@@MichelvanBiezen thank you very much professor🤩
@@MichelvanBiezen Hai prof, hope you doing well !
Yes, thank you. 🙂
how did you know there is upward force on pivot point 2?
Because when you pick the pivot point (the point of rotation), there are only 2 torques. The first caused by the weight of the person, the second by the support point. The sum of the torquest must be zero. When you move to the second pivot point (at the support point) you do the same. For the sum of the torques to be zero, the force must be downward.
@@MichelvanBiezen Thanks so much!
When you write out the net Y-force equation, how come we don't have to account for the normal force that the two supports exert onto the diving board?
Bryanna Phan The equation does include the normal force ----> F2
I thought F1 was also a normal force, which means that it should be pointing up. so we should have it as a positive force?? is that not the case?
It depends on the perspective. (Think about Newton's third law)
I'm confused, for d1 you have 3m but shouldn't it be 3m + 0.5m since the pivot isn't on the edge of the pool? Same with d3, I would think its 2m - 0.5m?
+DrDrewsAdventures d1 = 3 m = distance from the person to support 2
Question: why didn't you took the distantie to calculate F2 in the Fy direction? Thanks !
Greatz from Belgium
Ayoub. The distance is only needed when you calculate the torque. (Torque = force x distance).
When I calculated the sum of the forces in the y-direction only force is needed. Using Newton's second law in the y-direction (sum of the forces in the y-direction = 0 if there is no acceleration in the y-direction)
P.S. I used to live in Belgium
Thanks thanks thanks thanks. Really.
The sum of forces in the y direction should be 0. That means F2 = Weight due to board + Weight due to person - F1. NOT PLUS F1...So F2 should be 147N so that F1 + F2 = W Person + W Board...
The sum of the forces in the y-direction in the video are equal to zero.
If you add up the reactive forces or normal forces from the posts you don't get the force of the the weight from the person and the board.
Total downwards forces: (40*9.8) + (60* 9.8) = 980N.
Total upward forces: 735N + X
Total Down - Total Up = 0
Total Down = Total Up
980 = 735 + x
x = 245 --> F2 = 245N not 1715N
Either that or I must have made a mistake somewhere.
PS, thank you for your education videos. They are highly appreciated.
Fup = F down
1715N = 735 N + (100 x 9.8)N
(thanks for checking)
Sir in problem 2 how you take 1/5L???
Plzz answer me
Which problem 2 are you referring to? If you are referring to another problem, it would be better to place your comment there.
Thank you!!
So F1 has a downward force. By Newtons 3rd law isnt there a reacting force? A FNormal acting upwards as well?
What if there was a newton scale beneath that support beam, what would it read? How do you go abouts that? Im guessing 2 FNs that of the floor and that of the support on the beam.
F1 is the force exerted ON the diving board BY the bracket holding the diving board on the far left. The direction of F1 is downward. According to Newton's third law, there is an equal force | F1| pushing upward by the diving board pushing upward against the bracket. Since both forces are equal in magnitude and opposite in direction, there will be no acceleration (Newton's second law), and therefore, the diving board will stay in place.
OK thanks! :)
How is d1 =3m because 3m didn't reach the pivot point
Yes, the arrow didn't quite make it to the pivot point, but it is indeed 3 meters to the pivot point from the diver.
@@MichelvanBiezen Thank you
i don't understand why F1 is consider as counter clockwise ? isn't it in clockwise direction?
That depends on the reference point. In this case that is point 2. And therefore relative to point 2, F1 acting independently of any other forces would cause the diving board to rotate in a counter clockwise direction about point 2.
How about the normal force?
If you are referring to the normal force at the middle support, we don't need to take that into account since it does not produce a torque (when the point of rotation is taken at that support point)
@@MichelvanBiezen okay, sir. Thank you. Then, is there normal force exerted by pivot?
Yes, there is.
killer module
Sir how did you get 1/2 im confused how? Pls answer thank u
Are you asking about why d2 = 0.5 m ? Is yes, 0.5 m is the distance from the center of mass of the board and the pivot point. (the point where the board is supported)
Sir, What is mean by the BOARD STARTS TO TIP?
If not held down, the board would start rotating.
oh man, thanks! you don't know from how long I wanted to clear this concept of mine.
Thanks a lot.
And there is just one more thing of the board starts to tip then there will be no force on the board at first support at F1
That is correct, it is the force at point 1 that keeps the board from tipping.
Could we have decided to make mgd2 and F1d3 going counterclockwise? And mgd1 would turn then clockwise?
Liana,
The direction for each torque, , clockwise or counterclockwise, is determined by asking the question:"How would the beam turn about the chosen pivot point if that was the only torque acting on the beam.
Thus it is NOT arbitrary.
The convention however is that counterclockwise is a positive torque and clockwise is a negative torque.
I reversed the convention (clearly marked) to show that it doesn't matter which direction you make positive and which direction you make negative.
But the direction of the torque depends on which way the beam would turn if it was the only torque.
Michel van Biezen Thanks! Do you have videos on torque relating to pulleys using the inertia equation? (Where the system is not in equilibrium)
Liana Banana
Liana,
Take a look in the playlist:
PHYSICS 13 MOMENT OF INERTIA APPLICATIONS
Bedankt voor de snelle respons !
Ik vind het geweldig dat mensen van de hele wereld deze videos aanbekeikt. Tot hier toe studenten van meer dan 190 landen hebben deze videos gevonden.
Michel van Biezen Do you have Dutch relatives?
Laurelindo
Laurelindo,
Not Duch relatives but Belgian relatives. My native language is Flemish.
isn't F1*2.5 instead of 2?
Ali,
No 2m is correct.
why aren't we considering any normal forces here? even in the hungry bear video, there was a normal force on the bear that you didn't include.
+Shameera Rafeek We did not ignore them. There are two normal forces. The first one is at the pivot point, but because the moment are of that force is zero, since it is at the pivot point, it does not cause a torque. The second one is F1. That is the unknown force we had to calculate.
There is a normal force acting up on the person standing at the edge of the board...
so confused to as how u got d2=0.5m..from person to pivot = 3 and from f1 to pivot is 2......
The big M represents the mass of the diving board and thus d2 is the horizontal distance from the center of mass of the diving board to the pivot point (2), which is 1/2 meter.
@@MichelvanBiezen expound pls
Is this even right I mean two pivot points doesn't make sense
You can place the pivot point (or point of rotation) anywhere you like
should you include vector notation in these problems?
It depends on what you are trying to accomplish. If you are only after the magnitude of the force, then there is no need. There are plenty texts that do not include vector notation to keep it simple.
Torque 3 of 7 the answer is 44100N not the number you found on your calculator.
the F2 should be 735N because there is still a Normal force you didnt include the Normal 1 and Normal 2 the equation should be F2 = F1 because the Mg will be canceled out because of the normal force and also mg will be canceled out still because of the normal force
+John Raven Engalan
The answers are correct as is in the video.
ohhh ok thanks can i ask why is that in the summation of forces along y-axis why is there no normal force? i just want to clarify cause your videos really helped me passing my exams in physics
and I thought it was just d1 and d2. didn't see d3 coming
if i start with the first support , the force will be 735 and if i start with second support , the force will be 735 what is my mistake ???
net torque = zero
D1*F2-D2M2G-D3M3G
F=735
------------
D1*F1-D2M2G-D3M3G
F=735
Jana,
You need to clarify a few things. You indicate F=735 but you don't have F in either one of your two equations.
Second, place the numbers in the equation, so I can see how you calculated the results in each case. Without that I won't be able to tell.
am i the only one having problems with the audio of this and all the other 6 videos of this tutorial..??
Our older videos were not recorded in stereo. We have fixed the problem for our newer videos.
sir please help me out in this problem:- why didn't you consider the normal forces of the supports,diver and that of the board while calculating the net torque and does F2 account for all the reaction forces in the y direction and I think that F1 must be the weight of the support acting downwards
+Rahul Tiwari
The key to doing problems like this is to choose a pivot point at the location of one of the unknown reaction forces. (F * d = 0). That is why we didn't need to know F1 to solve the torque part of the problem.