Physics - Mechanics: Applications of Newton's Second Law (12.5 of 20) Corrected Video

This is a corrected video to a video that has an error ( • Physics - Mechanics: A... )
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In this video I will show you how to calculate the forces needed for a person pull himself up a stationary, moving, and accelerating elevator.
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[@srijahnavi3320]

I have always struggled in these types of problems, thanks a lot for clearing my doubts.

[@MichelvanBiezen]

Glad to hear that

[@RobertJohnson-nq2uy]

Helped me figure out a problem in the Feynman lectures, appreciate it!

[@ALAL-jm1lr]

hi, i think the trick to understand why T1 is always mg+Mg, even though there is new pulling(grasping) downward from the person inside elevator is to think as this: 1>idea1: what if the end of rope where person is catching is connected to same elevator . now total weight is divided between two ropes...and T1=mg+Mg. 2>idea 2: u may say hey..there one end of rope connected to elevator which have pulling force as total weight which is mg+MG, and there is new pulling force from the guy who catching the other end.... so T1 must be adding mg+Mg+guy pulling force.....but wait as this guy pull the rope....his weight mg inside elevator is reduced: as guy pulling force increase==>mg decrease of same amount.... and so total T1 is keep same as before mg+Mg.... hope this explain it more logical.....

[@imyoyoliu]

I was confused by this just now, Thank you!!!

[@trafalgardyobk7571]

If person pulls the rope then his wheight will be reduced to 0 and T1=50(9.8) and T2 =25(9.8)+[50(9.8)-25(9.8)] 25(9.8):means that wheight of man is transformed to T2 and [50(9.8)-25(9.8)]:means additional force that the man should add to equal the elevator weight Is this right?

[@MichelvanBiezen]

Did you get a different result compared to the video?

[@trafalgardyobk7571]

​@@MichelvanBiezen Thank you for your response Yes I got a deffirent result: If what I said is right then: T3=T2=490N And :T1=2(490)=2Mg=980N

[@MichelvanBiezen]

Which part of the problem are you referring to? For parts a, b, and c, T2 = T3 = 367.5 N And for part d, T1 must equal 810 N and T2 + T3 must equal T1 and T2 must equal T3

[@trafalgardyobk7571]

​@@MichelvanBiezen I'm talking about situation a : If the person pulls thé rope he will stick to the ceiling of the elevator therefore his weight will be transformed to the other Rope then: T2=Mg and T3=mg and he must pull very hardly to add additional force that will make T2=T3=Mg=mg+force of friction between man's hand and rope T3 then : T2=T3=MG=490N and T1=T2+T3=2MG=980N

[@ShimaaMohamed-jf9jl]

Thank you very much, we appreciate your effort to help us and explain the information in a simple and wonderful way

[@MichelvanBiezen]

You are welcome. Glad you found our viceos. 🙂

[@nikkireddinikki4418]

if the lift is moving down with acceleration 1 m/s2

[@michaelsilagy4659]

I'm still a little confused like the others in the comment section on how there is a mechanical advantage to this pulley system. From my understanding, if you were to hang an object weighing 37.5 kg (367.5N) on the other side instead of applying a pulling force of 367.5 N, the lift on the left side would start to fall, yes? Why does pulling on the rope then create a 2:1 difference in the needed force for there to be no acceleration? Thanks!

[@MichelvanBiezen]

That would not have the same effect. If a person in the box pulls down on the rope, it has the effect of reducing the effective weight by the same amount. Hanging a weight on the rope would not reduce the effective weight of the box.

[@gkx9914]

@@MichelvanBiezen i love u man

[@kartikeyrajgupta7191]

Hello sir I am Indian, your explanation is very clear Thank you so much sir

[@MichelvanBiezen]

Welcome to the channel

[@shoaibqureshi9818]

Hi sir what if the person leaves the rope then the lift will go free fall but will the man will go up or down after leaving the rope

[@MichelvanBiezen]

After the person lets go of the rope, both the elevator and the person will experience free fall and the person will sense to be weightless.

[@shoaibqureshi9818]

@@MichelvanBiezen isn't the normal reaction on the head of the man in upper direction applied by the lift when he was in equilibrium

[@MichelvanBiezen]

The head of the person was not touching anything.

[@dhruvmk3055]

I have a question. Let's analyze just the boy and the carrier. There are too forces acting on them - force due to gravity and force tension from the rope. Since they are in static equilibrium, the net force has to be zero. Thus, Tension must be equal to force due to gravity, which is (M+m)g = 735 N. Why is this wrong? Seems like perfectly logical physics to me.

[@MichelvanBiezen]

There are 3 forces acting on the boy and the box

[@dhruvmk3055]

@@MichelvanBiezen What is the third?

[@MichelvanBiezen]

The reaction force to the pull of the boy.

[@raven4234]

You are right that the upward force on the boy + carrier is from tension, therefore the total contribution from tension forces must be 735N. However, that total upward contribution from tension comes from two ropes. Because it's equally distributed between the two ropes (because of the pulley), then each rope accounts for 367.5N of that total force of tension. Not trying to discount the instructor's answer -- I think this is the same as what he said, just using different words.

[@ashwina5472]

sir,if we put a weighing machine of weight 1kg where the man stands will it show the weight of man as 12kg

[@ashwina5472]

Tension =(M1+m2+M3)g/2 Normal force of man=M1g-T is this solution correct. M3=mass of weighing machine

[@MichelvanBiezen]

Close, it would be 12.5 kg.

[@ashwina5472]

sir if the weighing machine itself has a mass of 1kg wouldn't it be 12kg

[@MichelvanBiezen]

Yes, including the mass of the scale it would be 12 kg.

[@ashwina5472]

thank you sir

[@ustka1]

Good day Sir, I don't understand why the school teaches that it is possible to pull yourself with this kind of solution. It takes 3/4 ths of the power to pull oneself with this basket in comparation to pulling yourself up.

[@pipertripp]

I don't understand how this pulley system is giving a 2:1 advantage (around 3:45). The weight of the lift + person is 735N and that's directly supported by the rope so that rope must have a tension of 735N, no? The force down should be 735N since this is a 1:1 MA setup, right? What am I missing here? See example 1 in this other video of yours: kzbin.info/www/bejne/bHjYnZevqKiJoKs

[@MichelvanBiezen]

If the person pulls with a force of 2 N then the tension holding the lift will now be 733 N. If he pulls with a force of 367.5 N the tension holding the lift will be 367.5 N

[@THEINNERCIRCLECROATIA]

The tension in the rope is the full weight of the person and the elevator since there is a force pulling on each side of the rope he isn’t explaining this well but the external force acting on the rope is 1/2 mg in a perfect word with a lot of assumptions

[@y.st.b5843]

2:42 I think it's T2 NOT T1 ??

[@iyadindia862]

Any body here how the case 2.25 is different from normal case in which if a mass is suspended over pulley attached with one end of rope over the pulley,to lift the object we have to apply a force equivalent to its weight Any body here pls make clarification

[@scimaniac]

What would happen if the person pulled more than or equal to his body weight?

[@MichelvanBiezen]

The acceleration would be greater, off course, it would not be possible to pull with a force greater than your body weight.

[@scimaniac]

Can you point me to a video explaining why I can’t pull more than my body weight? Equations will be helpful.

[@MichelvanBiezen]

If you hang from a rope what is the tension in the rope? (answer: mg which is your weight) It is not possible to pull down on a rope with a force greater than your weight.

[@scimaniac]

Thanks for the clarification. I did not think like that. I thought the person could not pull with his hands with his own weight. I agree that he can’t hang more than his weight;however, can he pull with his hands his own weight or more while in the cart not hanging.

[@krishnakumar-gw9vd]

If 2 pully and lifts by both hands

[@MichelvanBiezen]

It depends on how the pulleys are connected. We have a number of examples of multiple pulley problems.

[@fatimahrashid2966]

sir, in case (A) shouldnt the total tension be equal to 2*735 N since the the system is stationary and must therefore have equal weights suspended from both the rope sides? thanks in advance!

[@MichelvanBiezen]

Take a look at this video, it may help you understand this tricky concept: Physics - Mechanics: Newton's Laws of Motion (17 of 20) Third Law: Example 5

[@fatimahrashid2966]

Michel van Biezen thank you so much!

[@fatimahrashid2966]

Michel van Biezen in case of the 2 men pulling on a string from 2 sides, if the forces are unequal, say 10 N and 30 N, will the spring scale show a tension of 10 N?

[@MichelvanBiezen]

If the forces are not equal, there will then be a net force = (30N - 10N) = 20 N in the direction of the 30 N force and the system will correspondingly accelerate according to Newton's 2nd law: Fnet = m a

[@danieljonatan5182]

im still confused , is the work the same ? because if we assume that the person goes up 1 meter. the work = (T2+T3) 1 meter. On the other side the hand from the person =T3 * 1 meter ??? Thanks

[@danieljonatan5182]

why is there a mechanical advantage ?

[@MichelvanBiezen]

Sometimes in the case of a simple pulley the advantage is in the direction of the force applied. With a pulley you can lift heavy objects by using your weight as opposed to pushing the weight of the object up directly.

[@danieljonatan5182]

Thankyou sir for your answer, but i mean , that in this case, if we imagine that there is a coordinate. This person will go up 1 Meter with the force(T2+T3) and the rope will go down 1 meter with T2. Is this possible ?

[@nicholaslombardo2775]

All the answers should be doubled as you could see expecially towards the end of the video he writes the wrong subscript causing all the answers to be scaled down by a factor of two.

[@MichelvanBiezen]

The answers are correct in the video.

[@muddassir1975]

What is the force that the person exert on the lift?

[@MichelvanBiezen]

The force the person exerts on the floor of the lift = Mg - F

[@iyadindia862]

My question is this How its possible to set a body into uniform motion from rest without applying any additional force Or what makes the rest lift into uniform motion Pls clarify

[@samarpitabhattacharya5953]

The term will be external force and to bring a body to any kind of motion you need to apply external force on it otherwise it's not possible..... so first you apply some external force on the body at rest, the force accelarates the body then apply a nonuniform retarding force on it until it reaches your desired uniform velocity and you are done

[@devyn003]

is this not wrong? In this video it should be. Static: Fnet = 0 (Force of the box)+T2=0 T2=g(75kg) = T3 T1=T2+T3 = 2g(75kg) I feel like he forgot which was T1 and rolled with it.

[@MichelvanBiezen]

T1 is correct in the video.

[@sidharthsamal7418]

Tysm from India

[@MichelvanBiezen]

Welcome to the channel.

[@adhritproductions]

Did he just use calculator at 3:35 to divide by 2?

[@priyankakhanagaonkar2464]

They are different from Indian

[@petereziagor4604]

Haha, one way I understood this problem is to see the whole weight as one. If the person is not applying the force, the whole system would fall at g and to stop it from falling, the hang has to take up the tension of the whole weight and since it's in a pulley system, the system has to be balanced. Of course, the person can't apply a force greater than his weight and the tension on the strings pulling on the hang must be equal, pulley system.

[@MichelvanBiezen]

Good insight.

[@shanmukhm1295]

FBD IS WRONG...APPRECIATE IF YOU .CORRECT IT....

[@ahaanverma1027]

god bless

[@gxiong3917]

Please give the answer before the explaination.