I took physics for engineers in college and it was one of hardest class I've ever taken. Out in the real-world now, we're having an issue with one of our products and I believe it to be caused by using a spring from a different manufacturer. About to do this experiment to find the K values of both springs and see if there is a difference. Pretty exciting when it's real world applications lol. Not so much on paper with 30 other exercises to do.
What a fantastic video! Thank you so much for uploading. Cannot imagine how you wrote everything backward behind that transparent board though. It must be very challenging.
Your way of presenting this on glass & explaining it is a great way. Would like to see this method used in schools. I sincerely hope you are a teacher !! Peace.
He is writing using his Left Hand. This might be so if He is a Right handed then he did it all the calculation facing towards himself while shooting and during post production it was switched to Mirror Image (Lateral inversion was achieved/flipped). Thus giving you an impression that he is writing in reverse ...
This video is beautiful.My question is why can't one use Ls, I mean when it's stretched it becomes the new length right? And do we always assume that the spring is in equilibrium? Because my question says "Spring is suspended in a vertical position" do I than assume that the sum of forces in the y direction equal to zero? I hope I was clear
Hi Lee-Roy, thanks for the feedback. I'm not sure what you meant by your question. We are definitely using Ls, but only in conjunction with Li. The key factor in spring problems is "how far does it stretch from its equilibrium length" which means we want to use x = Ls - Li. For your second question, if the hanging mass is stationary (hence not accelerating), then the sum of the forces has to equal zero. Hope this helps. Cheers, Dr. a
I have a question, If I were measuring in cm and the mass was in grams, for when when I do my calculation of k = mg/Ls-Li. Would I convert g which is 9.8m/s^2 to 980cm/s^2?
1113 Himanshu Thakkar, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
X is the difference between the pointer reader and the initial reading, right? What if we have more than one pointer reading. We had five loads with a common difference of 10g. How then do we get 'X'. I need help, please.
The Shirt pockets are Missing and the Buttons are their for you to Make out that he is writing with the Right Hand and The Video was laterally inverted after shoot during Post production ... He is a right handed person ...
Does the spring constant stay the same if you increase the unscratched length of the spring (assuming this spring is of the same properties as the last but shorter one)?
Nicholas Haynes, Yes in theory. No in practice. Think of stretching the spring until the coiled wire is completely straight. Then it clearly won't act like a spring anymore. All springs will eventually behave nonlinearly, that is the restoring force is not proportional to stretch. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
For a helical spring with a circular wire, the formula that gives the spring constant in terms of its geometry is: k = G*d^4/(64*n*R^3) Where: G is the shear modulus of the material, a measure of its rigidity. d is the wire diameter n is the number of coils R is the radius of the centerline of the coils as they are wrapped around the helix. Two otherwise-identical springs, which are each initially built with the same number of coils, but different lengths, should have the same spring constant.
Yes, absolutely. But keep in mind that if you stretch it far enough (as with any spring), it will no longer be linear: the force will not be proportional to displacement. Cheers, Dr. A
What would happen to a spring if you continue adding weights on to the spring, and why would it be unsafe. Is it because it has extended its elastic limit, and it could snap. Thank you :)
Yes, absolutely. A spring is, after all, a piece of coiled wire. Stretch it far enough and the wire straightens out and no longer acts like a spring. Stretch it beyond its threshold limit, and it will snap. Cheers, Dr. A
What if the mass is placed on the spring and it stretched it down 31cm, and then it is pulled down by someone 3cm more. Would the length of the spring be the stretch of the mass and then the initial length be 3cm?
The x-value for the equation F=k*x, would be 34 cm in this example. The force in the spring would be resisting the weight of the mass, plus the human force applied to it.
When k=mg/x and when k=2mg/x?? If string is stretched just by weight, where does the gravitational potential energy goes if only half is converted to elastic potential energy? I'm hella confused..Somebody help:(
Very nice video sir... Sir I have a question.. Say we have two blocks of equal masses connected to the two ends of a spring of spring constant K and the whole system is kept on a smooth horizontal frictionless table. Now we apply two equal pulling forces on both the masses and as a result both the masses get displaced by (x) metre say.. Naturally the spring will extend by 2x metre. Now we apply the energy conservation principle Elastic potential energy stored in the spring = Net amount of work done on both the blocks Now, Elastic potential energy stored in the spring = 1/2*k*(2x) ^2 = 2kx^2 Net amount of work done on both the blocks = 1/2*k*x^2+1/2*k*x^2 = kx^2 But these two values are not equal Sir please explain why??
Excellent question! Your first calculation is correct but the second one is not. Imagine this: you pull one block out a distance x. That amount of work is certainly 1/2*k*x^2. The "official" way to do this is to integrate the force (kx) from 0 to x. You integrated from zero because that was the rest length. The spring is not pulling on the block at this position. However, when you now pull out the second block, you have to integrate from x to 2x, since the block is initially being pulled on by the spring. When you do this properly, you end up with 1/2*k**(3x)^2. When you add this to the first answer, you get 2kx^2. Cheers, Dr. A
But when we are applying two opposite forces at the same time, then for both the forces isn't the spring starts getting stretched from its natural length x=0 ?
Yes, but the force is bigger than kx in that case. You are effectively cutting the spring in half (center stays fixed), and that means that k doubles (a short spring is stiffer than a long spring). So that's where you get your factor of two. Very good questions, though. Cheers, Dr. A
Zamaswazi, We usually put the negative sign in front of g. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
good day.. I have a question.. I want to design a pogo stick that can hold a max. weight of 90kg that reaches a height of 0.6m. is the K calculated the same way?
+Rae Clarke Hi Rae, good question! No, you'll have to get a little more complicated with your calculation. The important factor, other than K, is how far can the spring compress without "bottoming out." You will have to use conservation of energy to drop a mass from height h, compress the spring a distance x, and make sure that this distance x is still within the working distance of the spring. Give it a shot and let me know how it turns out. Cheers, Dr. A
if solving for x, given a mass (2kg) and only Li (8cm), do you find the spring constant by k = 2 x 9.8 / 8 then find x = 2 x 9.8 / k and thus Lstretch Ls = 8 + x OR is it k = (weight of Li) x 9.8 / 8 then x = 2 x 9.8 / k etc?? it seems like the latter but my math skills are rusty
Cynthia, Not quite. Remember that Li (8cm) is the rest length. That is, the length of the spring with no mass hanging on it. When we hang a mass of 2 kg it stretch the spring a distance x = 4 cm. See the discussion at the 2:40 mark in the video. (To calculate k in your equation, change your 8 cm to 0.04 m and you're good). Best of luck, and keep up with the physics! Cheers, Dr. A
thank you Dr Anderson so much for the response! let me ask this another way can you solve for k on a spring at rest, only knowing Li and mass? or is there a another equation to find x, only knowing Li and mass?
Since at rest kx=mg, there are three unknowns (k,x,m). You need to know two of these to get the other one with this equation. Rest length Li doesn't help you, because the important quantity is x: how far does it stretch from the rest length. Keep working with it. Cheers, Dr. A
Dr Anderson, thank you for helping me see (k) is only determined after measuring (x) thus true (x) can only be determined by field test, then math can determine changes with different (mg)s thereafter I was seeing since (k) is inherent it could be determined at rest (maybe it could but perhaps the spring's area, helix diameter and/or pitch would have to be formulated? just not by Hooke's) thanks again
Hello, I did this experiment with my students wit a 15 N/m spring and we found after a least square fit that mass= 1,6275xdisplacement + 0,0251. The measurement was very acurate. We found that the spring started to strech with 25 grams. I recomended the student to use the slope in this case 1.6275 kg/m and multiply this by 9.8. This is a real spring and does not uses the idea of k=F/delta but k=deltaF/deltax. Please comment
+Ignacio Olivares Ignacio, love it! This is great, and you are absolutely right that this is a better way to determine k. Since F = kx = mg, if we include a Δ on each side of the equation, we get kΔx = gΔm and therefore k = g(Δm/Δx). So all you have to do is add a bit more weight (Δm) and measure how far the spring stretched (Δx) and you can calculate k from the above equation. Cheers, Dr. A
+Matt Anderson that's fine. The best is making deltas on both sides of the equations. When students are doing the experiment it is useful to plot the data first in order they see the need to do kΔx = gΔm. This became now a very useful formula to measure spring constants. Is your image inverted?
Sir, if I cut a spring of force constant k in 3 pieces into the lengths of ratio 1:2:3 ,then what will be the force constant of these springs? Explain please ,sir.
Good question. The short answer is that spring constants "add up in reciprocal" when you connect springs in series, and add up directly when you connect springs in parallel. So the answer is that the new spring constant becomes 3*k, when you cut a spring in 3 pieces. Here's why. Springs in series by definition, carry the same force (F) in each of them. Suppose we have identical 3 short springs, with a spring constant of c. Under load F, individually they would each deform a distance lowercase d. So for each individual spring, F = c*d. When we connect them in series, the deformations add up to a total deformation capital D. D = 3*d, because there are 3 springs deforming identically. Define as the spring constant of the three springs, such that F = k*D. Solve for k, k = F/D. Replace D with D=3*d, and get, k = F/(3*d) Replace F with F=c*d, and get k=c*d/(3*d). Result: k = c/3, or c = 3*k. The new spring constant after cutting a spring in thirds, is three times the original.
Yes. Then you would apply Earth's gravity 9.8 N/kg, to switch from mass in kg, to weight in Newtons. Mass = 50 grams = 0.05 kg Corresponding weight = 0.05 kg * 9.8 N/kg = 0.49 N deformation, suppose 2 cm, which is 0.02 m k = 0.49 N / 0.02 m = 24.5 N/m
This is a really helpful video, but how would I design an experiment to determine the spring constant for springs in series (attached end to end) springs in parallel? Any ideas?
+Beatrice Brown Great question. Springs add just like capacitors (and the opposite of resistors). Parallel: If you attach two springs in parallel, the effective new spring will be kp = k1 + k2. This means of course that you have made a "stiffer" spring (i.e. an object of weight mg will not stretch two springs as far, if both springs are attached directly to the object). Series: If you attach two spring in series, the effective new spring will be 1/ks = 1/k1 + 1/k2. This means of course that you have made a "weaker" spring (i.e. an object of weight mg will stretch two springs further, in total, than one spring). Hope this helps, and hopefully with this info you can design an experiment to test it out by hanging some masses from various combinations of springs. Cheers, Dr. A
Wire gauge is just a measurement of the wire diameter. And for historical reasons, a gauge with higher number is actually a smaller diameter. See en.wikipedia.org/wiki/Wire_gauge Cheers, Dr. A
Matt Anderson yes i understand that part you are telling me. But I was reading for some books when you are designing a spring. You have to choose a random wire diameter. My question was, how do you determine that diameter. I was reading that there are some security factor and things that assure the spring will work good
@@yoprofmatt yes lolol, the comment section be very shocked that you can write backwards sksksk. Thanks for your wonderful explanation tho. Helped me a lot
Simply: mg = mass times gravity which in that case is 2 kg times 9.8 m/s^2 which if we calculate equates to 19.6 So now we have 19.6/X To solve for X you have to convert to SI units (metres in this case) and since you have 2 sizes one for initial size and the second for the stretched size you have to get the size between them which you get by doing L2 - L1, L1 being 8 cm and L2 being 12 cm So now we got 12 - 8 which equals 4 cm or 0.04 metres. If we plug the 0.04 metres into the equation you get: 19.6/0.04 Which equates to 490.
a crate is hung from a spring with a force constant of 525 N/m. This stretches the spring 0.30 from its equilibrium position. What is the mass of the crate????????????????????????????????????????????????????????????????
Shadow Wind, Acceleration due to gravity. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Alpha Junior, Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass Cheers, Dr. A
![ประสบการณ์ จ้าง Developer ที่ผมจะไม่มีวันลืม [ภาค1]](http://i.ytimg.com/vi/bzivN4KDg-k/mqdefault.jpg)
I’m just amazed that this dude can write backwards so well
@Akshath bruh
i think he writes normally, and then flip the video over, so we can read it.
Edward Huang I have been studying physics for about 3 years now and I guarantee you I would have never guessed that
@Akshath no shit sherlock
This is a glass table.. Camera put opposite to him
I took physics for engineers in college and it was one of hardest class I've ever taken. Out in the real-world now, we're having an issue with one of our products and I believe it to be caused by using a spring from a different manufacturer. About to do this experiment to find the K values of both springs and see if there is a difference. Pretty exciting when it's real world applications lol. Not so much on paper with 30 other exercises to do.
How long did it take to learn to quickly to write backwards?
They mirror the footage
If it is mirror then the letters should also had to be in reverse direction .
He's writing on a glass panel......
Joey A he knows doubt anyone that dum lmao
I still have no idea
What a fantastic video! Thank you so much for uploading. Cannot imagine how you wrote everything backward behind that transparent board though. It must be very challenging.
+实在哈哈
Thanks 实在哈哈, but I didn't write everything backwards (that's pretty hard). Check out the secret here: learningglasssolutions.com
Cheers,
Dr. A
The most helpful video I have found so far!
Thank you very much
Your way of presenting this on glass & explaining it is a great way. Would like to see this method used in schools. I sincerely hope you are a teacher !! Peace.
You are the man for me Sir.No messing about.Straight to the point.
Just trying to keep it honest.
Cheers,
Dr. A
This has melted my brain, but then again, other stuff melts my brain even more. Thanks for the simpler stuff!
Great Stuff! School project solved for me in 6:44 min. Now just remain to rewrite the video.
Thank you so much for this, it's really going to help me for tomorrow's final exam!
Same.. but 3 years later
@@tytheguy1771 😂 if you have an exam soon, I wish you the best of luck man.
@@vesperowl3626 thank you, i write my final tommorrow morning, i need as much luck as possible
@@tytheguy1771 I wish you all the best man, tell me how it goes tomorrow ❤
@@vesperowl3626 88%! this vid definitely helped.
Thank you for making a easy to understand straight forward video! It was very helpful!
This is literally so clear and a life saver, tysm
He is writing using his Left Hand. This might be so if He is a Right handed then he did it all the calculation facing towards himself while shooting and during post production it was switched to Mirror Image (Lateral inversion was achieved/flipped). Thus giving you an impression that he is writing in reverse ...
Wonderful lecturer that makes physics fun and understandable !!!
Thank you for everything you’ve done for me
Nicholas Tovbin,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
This taught me better than my physics 201 professor, thank you so much
Thank you very much sir, it is extremely simple to understand for a beginner also.
DESIGN CENTRE,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
Thank you so much Mr. Anderson, you explained this very well and really helped me understand.
thank you so much I finally understand how this works
Awesome. Re-explain it to me in a few years.
Cheers,
Dr. A
finally a good example thank you so much
This video is beautiful.My question is why can't one use Ls, I mean when it's stretched it becomes the new length right? And do we always assume that the spring is in equilibrium? Because my question says "Spring is suspended in a vertical position" do I than assume that the sum of forces in the y direction equal to zero? I hope I was clear
Hi Lee-Roy, thanks for the feedback. I'm not sure what you meant by your question. We are definitely using Ls, but only in conjunction with Li. The key factor in spring problems is "how far does it stretch from its equilibrium length" which means we want to use x = Ls - Li.
For your second question, if the hanging mass is stationary (hence not accelerating), then the sum of the forces has to equal zero.
Hope this helps.
Cheers,
Dr. a
Great video Proffessor! It is going to help me a lot when it comes time for my Physics summative assessment tommorow!
+Victor Foresti Castro
Thanks Victor, good luck on your test!
Cheers, Dr. A
@@yoprofmatt .... hlo plz make a vedio on pulley's
Very easily explained...Good work
lots of love from Bangladesh
Thanks Im in Statics and forgot this LOL helped a lot
I have a question, If I were measuring in cm and the mass was in grams, for when when I do my calculation of k = mg/Ls-Li. Would I convert g which is 9.8m/s^2 to 980cm/s^2?
You are just awesome and mindblowing
1113 Himanshu Thakkar,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
could you just use extension = force/constant to work out the extension of the spring with the new weight ?
very good method, I really like this
Thanks. Enjoy the physics.
Cheers,
Dr. A
Thanks, glad to be of help.
Cheers,
Dr. A
Thank you, this really helped me :)
You are welcome. Keep up with the physics.
Cheers,
Dr. A
Clear as Mud lol... 👍 Genius Sir....I don’t know how I finish college 😅
X is the difference between the pointer reader and the initial reading, right? What if we have more than one pointer reading. We had five loads with a common difference of 10g. How then do we get 'X'. I need help, please.
this dude just saved my ass for the physics quiz
Awesome! Hope you nailed it.
Cheers,
Dr. A
@@yoprofmatt i sure did
The Shirt pockets are Missing and the Buttons are their for you to Make out that he is writing with the Right Hand and The Video was laterally inverted after shoot during Post production ... He is a right handed person ...
Correct!
The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
It was very clear! Thank you!
Thank you for the comment. Keep up with the physics.
Cheers,
Dr. A
it was really helpful.
thanks for hooking a brother up
Glad to be of help.
Cheers,
Dr. A
Thank you so much
Thank you so much. This was amazing.
Thanks man really helped a lot
+Abdullatif Alhor
Abdullatif, glad you found it helpful. Stay tuned for more.
Cheers, Dr. A
Interesting, well explained.
thank you for posting this. soooooo helpful
Does the spring constant stay the same if you increase the unscratched length of the spring (assuming this spring is of the same properties as the last but shorter one)?
Nicholas Haynes,
Yes in theory. No in practice. Think of stretching the spring until the coiled wire is completely straight. Then it clearly won't act like a spring anymore. All springs will eventually behave nonlinearly, that is the restoring force is not proportional to stretch.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
For a helical spring with a circular wire, the formula that gives the spring constant in terms of its geometry is:
k = G*d^4/(64*n*R^3)
Where:
G is the shear modulus of the material, a measure of its rigidity.
d is the wire diameter
n is the number of coils
R is the radius of the centerline of the coils as they are wrapped around the helix.
Two otherwise-identical springs, which are each initially built with the same number of coils, but different lengths, should have the same spring constant.
thank you this was very helpful
Emmett,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
Does this apply on a bungee cord? A small one, lets assume I'm using a bungee cord on a catapult, can I apply this to find it's constant? Thanks.
Yes, absolutely. But keep in mind that if you stretch it far enough (as with any spring), it will no longer be linear: the force will not be proportional to displacement.
Cheers,
Dr. A
in simple terms its just: constant is equal to mass multiplied by gravity divided by extension
What would happen to a spring if you continue adding weights on to the spring, and why would it be unsafe. Is it because it has extended its elastic limit, and it could snap. Thank you :)
Yes, absolutely. A spring is, after all, a piece of coiled wire. Stretch it far enough and the wire straightens out and no longer acts like a spring. Stretch it beyond its threshold limit, and it will snap.
Cheers,
Dr. A
What if the mass is placed on the spring and it stretched it down 31cm, and then it is pulled down by someone 3cm more. Would the length of the spring be the stretch of the mass and then the initial length be 3cm?
The x-value for the equation F=k*x, would be 34 cm in this example. The force in the spring would be resisting the weight of the mass, plus the human force applied to it.
When k=mg/x and when k=2mg/x??
If string is stretched just by weight, where does the gravitational potential energy goes if only half is converted to elastic potential energy? I'm hella confused..Somebody help:(
Very nice video sir...
Sir I have a question..
Say we have two blocks of equal masses connected to the two ends of a spring of spring constant K and the whole system is kept on a smooth horizontal frictionless table. Now we apply two equal pulling forces on both the masses and as a result both the masses get displaced by (x) metre say.. Naturally the spring will extend by 2x metre.
Now we apply the energy conservation principle
Elastic potential energy stored in the spring = Net amount of work done on both the blocks
Now,
Elastic potential energy stored in the spring = 1/2*k*(2x) ^2 = 2kx^2
Net amount of work done on both the blocks = 1/2*k*x^2+1/2*k*x^2
= kx^2
But these two values are not equal
Sir please explain why??
Excellent question! Your first calculation is correct but the second one is not. Imagine this: you pull one block out a distance x. That amount of work is certainly 1/2*k*x^2. The "official" way to do this is to integrate the force (kx) from 0 to x. You integrated from zero because that was the rest length. The spring is not pulling on the block at this position. However, when you now pull out the second block, you have to integrate from x to 2x, since the block is initially being pulled on by the spring. When you do this properly, you end up with 1/2*k**(3x)^2. When you add this to the first answer, you get 2kx^2.
Cheers,
Dr. A
But when we are applying two opposite forces at the same time, then for both the forces isn't the spring starts getting stretched from its natural length x=0 ?
Yes, but the force is bigger than kx in that case. You are effectively cutting the spring in half (center stays fixed), and that means that k doubles (a short spring is stiffer than a long spring). So that's where you get your factor of two.
Very good questions, though.
Cheers,
Dr. A
Thanks but may I ask why is g not negative 9.8 … I got confused
Zamaswazi,
We usually put the negative sign in front of g.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
good day.. I have a question.. I want to design a pogo stick that can hold a max. weight of 90kg that reaches a height of 0.6m. is the K calculated the same way?
+Rae Clarke
Hi Rae, good question! No, you'll have to get a little more complicated with your calculation. The important factor, other than K, is how far can the spring compress without "bottoming out." You will have to use conservation of energy to drop a mass from height h, compress the spring a distance x, and make sure that this distance x is still within the working distance of the spring. Give it a shot and let me know how it turns out.
Cheers, Dr. A
+Matt Anderson Oh ok cool.. no prob I will.. thanks
+Rae Clarke
Rae, just saw this video: th-cam.com/video/fyxoW21J7M8/w-d-xo.html
Thought you might be interested.
Cheers, Dr. A
Thank you, so much help
Great to hear. Thanks for the feedback.
Cheers,
Dr. A
Thanks 😊
if solving for x, given a mass (2kg) and only Li (8cm), do you find the spring constant by
k = 2 x 9.8 / 8
then find
x = 2 x 9.8 / k
and thus Lstretch
Ls = 8 + x
OR is it
k = (weight of Li) x 9.8 / 8
then
x = 2 x 9.8 / k etc??
it seems like the latter but my math skills are rusty
Cynthia,
Not quite. Remember that Li (8cm) is the rest length. That is, the length of the spring with no mass hanging on it. When we hang a mass of 2 kg it stretch the spring a distance x = 4 cm. See the discussion at the 2:40 mark in the video. (To calculate k in your equation, change your 8 cm to 0.04 m and you're good). Best of luck, and keep up with the physics!
Cheers,
Dr. A
thank you Dr Anderson so much for the response!
let me ask this another way
can you solve for k on a spring at rest, only knowing Li and mass?
or is there a another equation to find x, only knowing Li and mass?
Since at rest kx=mg, there are three unknowns (k,x,m). You need to know two of these to get the other one with this equation. Rest length Li doesn't help you, because the important quantity is x: how far does it stretch from the rest length.
Keep working with it.
Cheers,
Dr. A
Dr Anderson, thank you for helping me see (k) is only determined after measuring (x) thus true (x) can only be determined by field test, then math can determine changes with different (mg)s thereafter
I was seeing since (k) is inherent it could be determined at rest
(maybe it could but perhaps the spring's area, helix diameter and/or pitch would have to be formulated? just not by Hooke's)
thanks again
Hello, I did this experiment with my students wit a 15 N/m spring and we found after a least square fit that mass= 1,6275xdisplacement + 0,0251. The measurement was very acurate. We found that the spring started to strech with 25 grams. I recomended the student to use the slope in this case 1.6275 kg/m and multiply this by 9.8. This is a real spring and does not uses the idea of k=F/delta but k=deltaF/deltax. Please comment
+Ignacio Olivares
Ignacio, love it! This is great, and you are absolutely right that this is a better way to determine k.
Since F = kx = mg, if we include a Δ on each side of the equation, we get kΔx = gΔm and therefore k = g(Δm/Δx). So all you have to do is add a bit more weight (Δm) and measure how far the spring stretched (Δx) and you can calculate k from the above equation.
Cheers,
Dr. A
+Matt Anderson that's fine. The best is making deltas on both sides of the equations. When students are doing the experiment it is useful to plot the data first in order they see the need to do kΔx = gΔm. This became now a very useful formula to measure spring constants. Is your image inverted?
+Ignacio Olivares Dr.I.
+Ignacio Olivares
Indeed it is inverted. See this: th-cam.com/video/CWHMtSNKxYA/w-d-xo.html
Cheers,
Dr. A
If we place a spring on ground then
How we can calculate the spring constant?
Sir, if I cut a spring of force constant k in 3 pieces into the lengths of ratio 1:2:3 ,then what will be the force constant of these springs? Explain please ,sir.
Good question. The short answer is that spring constants "add up in reciprocal" when you connect springs in series, and add up directly when you connect springs in parallel. So the answer is that the new spring constant becomes 3*k, when you cut a spring in 3 pieces.
Here's why. Springs in series by definition, carry the same force (F) in each of them. Suppose we have identical 3 short springs, with a spring constant of c. Under load F, individually they would each deform a distance lowercase d. So for each individual spring, F = c*d.
When we connect them in series, the deformations add up to a total deformation capital D. D = 3*d, because there are 3 springs deforming identically. Define as the spring constant of the three springs, such that F = k*D. Solve for k, k = F/D. Replace D with D=3*d, and get, k = F/(3*d) Replace F with F=c*d, and get k=c*d/(3*d). Result: k = c/3, or c = 3*k. The new spring constant after cutting a spring in thirds, is three times the original.
if I hang a 50g weight on a spring I measure the deformation in cm, to find k I would put it into metres, would I change the weight from g to kg?
Yes. Then you would apply Earth's gravity 9.8 N/kg, to switch from mass in kg, to weight in Newtons.
Mass = 50 grams = 0.05 kg
Corresponding weight = 0.05 kg * 9.8 N/kg = 0.49 N
deformation, suppose 2 cm, which is 0.02 m
k = 0.49 N / 0.02 m = 24.5 N/m
Inspiring
Benson,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
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Dr. A
This is a really helpful video, but how would I design an experiment to determine the spring constant for springs in series (attached end to end) springs in parallel? Any ideas?
+Beatrice Brown
Great question. Springs add just like capacitors (and the opposite of resistors).
Parallel: If you attach two springs in parallel, the effective new spring will be kp = k1 + k2. This means of course that you have made a "stiffer" spring (i.e. an object of weight mg will not stretch two springs as far, if both springs are attached directly to the object).
Series: If you attach two spring in series, the effective new spring will be 1/ks = 1/k1 + 1/k2. This means of course that you have made a "weaker" spring (i.e. an object of weight mg will stretch two springs further, in total, than one spring).
Hope this helps, and hopefully with this info you can design an experiment to test it out by hanging some masses from various combinations of springs.
Cheers, Dr. A
thanks :) I've designed an experiment which we have to carry out this week, thanks again for your help :)
Over 500 companies in South-East Asia apply the patents of Veljko Milkovic's two-stage oscillator in their production
Tesla Tesla,
Thanks for the comment, and keep up with the physics!
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Dr. A
so helpful thank you
Thank you
Bevan A,
You're very welcome. Glad you're enjoying the videos.
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Dr. A
Hello
And before you find the K.
How can you determine the wire gauge.
Wire gauge is just a measurement of the wire diameter. And for historical reasons, a gauge with higher number is actually a smaller diameter. See en.wikipedia.org/wiki/Wire_gauge
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Dr. A
Matt Anderson yes i understand that part you are telling me.
But I was reading for some books when you are designing a spring. You have to choose a random wire diameter.
My question was, how do you determine that diameter. I was reading that there are some security factor and things that assure the spring will work good
You should consult the book "Machinery's Handbook."
Cheers,
Dr. A
nice video. thanks....but the glass board gives me a headache..
Is it that hard to guess that the video footage is actually mirrored?
IDK I’m broke,
No, not that hard. The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
@@yoprofmatt yes lolol, the comment section be very shocked that you can write backwards sksksk. Thanks for your wonderful explanation tho. Helped me a lot
how did you get 490? i just dont understand. can anyone tell me?
Simply:
mg = mass times gravity which in that case is 2 kg times 9.8 m/s^2
which if we calculate equates to 19.6
So now we have 19.6/X
To solve for X you have to convert to SI units (metres in this case) and since you have 2 sizes one for initial size and the second for the stretched size you have to get the size between them which you get by doing L2 - L1, L1 being 8 cm and L2 being 12 cm
So now we got 12 - 8 which equals 4 cm or 0.04 metres.
If we plug the 0.04 metres into the equation you get:
19.6/0.04
Which equates to 490.
@@Naeromusic where did he get the 9.8?
@@MyCrazy4life You should read your physics book. It's the gravitational acceleration of the Earth. 9.8 meters per second squared (^2)
matt, you didn't mention the initial tension which most spring have.
In civil engineering i found Spring constant Cfi kNm/rad, who determine this spring constant?
That sounds like a torsional spring constant, rather than a linear spring constant.
Hello Dr. Strange
i cant understand ??? which country you are ?
USA.
Cheers,
Dr. A
hey that rhymes
a crate is hung from a spring with a force constant of 525 N/m. This stretches the spring 0.30 from its equilibrium position. What is the mass of the crate????????????????????????????????????????????????????????????????
cool board! Where can I get one? ^_^
Your majesty,
Boards are available here: www.learning.glass
(so are build instructions)
Cheers,
Dr. A
haha, you are the first to call me "Your majesty" 😅
I hope springs are included =P
William Scally that was punny. xD
www.learning.glass
Cheers,
Dr. A
where did that 9.8 come from ???
That's the gravitational acceleration at the surface of the Earth.
Cheers,
Dr A
Yeah bhi nahi pata toh dekh kyoun raha hai
Im a student sir you have your calculater at graded it shoud be at degraded becouse a my answer was 163.24 and not 490
Does he have some sort of psychic powers? How on Earth can he write all of that backwards!?!?
Thank u
U welcome.
Dr. A
You make my physics book a waste of the paper it was printed on.
Thank u sir
You are welcome. Keep up with the physics!
Cheers,
Dr. A
why isn't mg negative?
It is because when you resolve, you get kx-mg=0, assuming you take up as positive. The minus is there.
Where did he get 9.8?????
Shadow Wind,
Acceleration due to gravity.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
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Dr. A
மிக மிக நன்று
how did he get 9.8??...help
The acceleration due to gravity near the Earth's surface is g = 9.8 m/s^2.
Hope this helps.
Cheers,
Dr. A
That's just the acceleration due to gravity at earth's surface, its a value you should memorize.
9.8 is the force of gravity on any object on earth. 9.8 m/s^2
Mr Fantastic?
Doubt it. But I'll accept Mr. Anderson (but only if you say it in the cool "Matrix" way).
Cheers,
Dr. A
Mr. Annderrrsonnnn
Excellent. I can hear the Matrix tone coming through.
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Dr. A
THIS GUY DRAWS BACKWARDS WITH HIS LEFT HAND WHAT THE SHLUCK
Alpha Junior,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
@matt anderson, omai
thanks
function
You look like gavin newsom's brother
Okay, cool. Can I use the governor's mansion?
Cheers,
Dr. A
can you do disney logo
Anyone ever nickname u banner?
Dr Strange?
It blows my mind how you write backwards.
Mike Riddle using a mirror may be
Mike Riddle just invert the camera during editing
he forgot to convert cm to m
yo hes writing backwards
bro the intro audio is way too loud jesus
what the hell
thank you so much
You're welcome!
Cheers,
Dr. A
Thank you
+muath jordan Muath, you are welcome. Keep up with the physics!
Cheers, Dr. A