Physics - Mechanics: Applications of Newton's Second Law (6 of 20) tension in elevator
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- Опубліковано 26 сер 2013
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In this video I will show you how to calculate the tension of a hanging 5kg object in an accelerating elevator.
Finally! A good tutorial. I've seen so many tension tutorials on UA-cam, and you make it SO much easier to understand. Thank you.
You're very welcome!
thanks a lot sir you just made physics more easier.keep up the good work!
Your teaching method is 100% worth
omg thank you so much sir, you made this topic so much easier for me! pls keep on helping others and i wish you all the best! thanks again sir
You are most welcome
I've been struggling with physics and this has helped so much thank you
Happy to help!
Thanks sir...
Thanks for you
thank you so much
Hi! This was really helpful! What do we do if there’s a second object attached to the first one?
We have a number of examples with Newton's laws applied to multiple masses connected, both horizontally and vertically in the playlists.
What an incredible explanation!!!
Thank you. Glad you like our videos. 🙂
how would you solve it if you dont have accelerations value but only velocity? Like 600kg and a velocity of 0.9m/s?
+legionostra
Set a = 0. Then T = mg = 49 N (for m = 600 kg --> T = mg = 600 kg * 9.8 m/sec^2 = 5880 N)
Thanks! But the acceleration should be a difference between up and down so, F=m(a+g)=m(m-g) and a=600g/1200+600
Hi again. T=mg +/- ma: I'm unsure if I am understanding this correctly; here's my thinking: I choose plus if the force applied to the object is in the same direction as the acceleration & minus if force applied to the object is in the opposite direction as the acceleration OR does it only have to do w the direction of gravity in all cases??? Thanks.
It's based on GRAVITY, right. I just watched the next video in the series.
For this type of problem (using Fnet = m * a), I find it easier to set Fnet = forcing aiding the acceleration (who are in the same direction) - forces opposing the acceleration (who are in the opposite direction).
Thank you. I'm watching all your videos...I eventually came across one where you explain this in detail and quite nice. It was a great help. I'd link it, but I forgot which one it was.
What if the elevators is moving down but accelerating up? Is it T=mg-ma?
It would be: T = mg + ma (you need the extra tension to slow down the elevator).
Thank you! Very well said
My pleasure!
We were taught earlier that g was negative, now here you have that it is positive. And if it were calculated as negative in this problem we would've found an answer of 39N and not 59N. Please clear this up, I'm starting to get very confused. Are you just taking absolute value to avoid negative force?
We are not trying to avoid anything here.
You are probably confusing acceleration in the y-direction (ay) and acceleration due to gravity (g). ay= -9.81m/s^2 g= 9.81 m/s^2
We normally use ay with kinematic problems because an object in free fall will always have an acceleration downwards, this is why it is negative.
When talking about forces we use the letter g instead of ay to indicate that the force of gravity is acting on the object.
These would give you the same value and sign: F = ma + m(ay) F = ma - mg
PD: ay = -g .....This is maybe why you got confused.
Always go with the direction of motion when you are picking directions for the other components, what I mean by that is when the whole "System" is moving in one direction, all components moving in the "System's" direction have a positive sign, and all components moving in an opposite direction have a negative sign!
In this example the motion for the whole system was going upward but the weight component is always moving downwards, hence it is put as negative.
Imagine if the elevator was accelerating downwards, then the motion for the whole system is going to be downwards resulting in the weight component being positive, and the tension being negative!
Hope this helps! Even though it's 3 months late haha :)
if mass is 1400kg and acceleration is 2.1m/s down what will my answer be? i used t = mg - ma but my answer marked it as wrong?
The answer would be 10,780 N Since the significant figures given equal 2, you may need to report the answer as: 1.1 x 10^4 N
thank a lot sir it was fast and easy
You're welcome!
Thank you!!!
You are welcome. 🙂
Sir, what if the lift is moving upwards at a constant deceleration or downwards at a constant deceleration?
What is the question? ("what if" doesn't reveal the question)
If the lift is going upwards with a constant deceleration, will the equation become
"T = mg + m(-)a" ?
thank you po
Could you please explain when do we use g as -9.8 N?
g is the acceleration due to gravity and it is directed downward.
Thus g = - 9.8 m/sec^2
Michel van Biezen I meant in the example you substituted g with (9.8 m/s^2) instead of (-9.8 m/s^2), but I think that this is related to the way you set the directions of your axis. Thank you so much :)
Khawla Mohd.
In kinematics the sign is important.
in other problems we just want the magnitude and the direction is determined in other ways.
Michel van Biezen I got it. Thanks for the clarification.
THANK YOOOOOU!
wouldn't the weight of the 5Kg block increase because to 59N as well because of its acceleration.
The actual weight will always be equal to mg = 49 N But the apparent weight (which is equal to the tension) will indeed be 49 N + 10 N = 59 N due to its acceleration
@@MichelvanBiezen so if we had an elevator going up at a certain acceleration, with a person inside of it. The person has an actual weight that never changes, but an apparent force/weight that is equal to the tension of the rope?
Yes. If you place a flat scale in the elevator and you then stand on the scale you apparent weight will increase as the elevator accelerates upward.
why isnt g negative if downwards is negative?
g is a vector quantity. If we use it as a vector, then the direction is negative. But if we use the magnitude of the vector it must be positive (since the magnitude of a vector can never be negative)
@@MichelvanBiezen thank you sp much!
This reminds me of Einstein's equivalence principle!
Arigato gozaimasu
You are welcome. Welcome to the channel! 🙂
@@MichelvanBiezen thx
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Fr
Santiago. I don't understand your comment.
@@MichelvanBiezen yea ok whitey
What is your purpose of making that an issue?