Physics - Mechanics: Applications of Newton's Second Law (16 of 20) pulley combination

[@jerrysmith9780]

I love that you did several examples, all very similar but slightly different and you repeated the methodology. Great reinforcement. Thank you so much. I learned a ton here.

[@mattthompson2385]

Sir, what do you do in the case if there's a double pulley system where one pulley is inside the other (the pulleys are attached)?

[@kasturivavilala6617]

it would be great if this can be extended to accelerating mass system. Can you plz help in writng the equations if the mass has an upward acceleration ,but both pulleys static?

[@carultch]

Suppose you want to make the mass accelerate upward at 1 m/s^2, and both pulleys and the rope are of negligible mass. The F=m*a equation on the mass is as follows: T5 - m*g = m*a Solve for T5: T5 = m*(g + a) From there, it is a matter of relating the tension T5 to the other tensions among the ropes. Because all the ropes and pulleys have negligible mass and friction, all tension forces acting on any segment of the system have to add up to zero, with the exception of the human force applying tension T1. T5 = T2 + T3, for force balance on the pulley by the mass. Forces have to add up to zero on objects of negligible mass, otherwise acceleration would be infinite. T3 = T1, because the upper pulley has negligible mass, and it just redirects the tension in the rope. Likewise, T2 = T3, for the same reason, for the lower pulley Put it all together, and we get: T1 = m*(g + a)/2 Plug in m=10 kg, a=1 m/s^2, g = 9.8 N/kg, and get T1 = 54 Newtons.

[@skp8444]

Very clear explanation❤

[@abruzzi24]

Thank you for all your vids doc!

[@tsepilembhele2418]

thank you...u r my lecture!!!!

[@cookiesncream8490]

what about the mass of pulley? does they effect the calculation or just ignore them?

[@MichelvanBiezen]

At the introductory level, we ignore the mass of the pulley, and in many real life applications, it is OK to ignore it. When we include the mass of the pulley, then you have to include the moment of inertia (see the playlists on that topic).

[@cookiesncream8490]

@@MichelvanBiezen alright sir, thank you very much.

[@binniethepooh2477]

This is so helpful 😘

[@hinsasimatupang323]

Thank you sir... you have helped me advance.

[@oghenemoses8839]

Sir, What is the relationship between T4 and T5 Since T4=T1+T2+T3 Hence I think T4= T1+T2+T3+T5 The sum of all the pulley attached to a dominant T4.

[@MichelvanBiezen]

T5 = T2 + T3 Therefore T4 = T5 + T1

[@oghenemoses8839]

@@MichelvanBiezen Thanks for your response, sir.

[@MrCoccimiglio]

This helped with my homework

[@ilyassourahou7839]

sir for the pulley their is a weight(mg)and it has an effect on it so it will change the value of T4 please sir could you explain it ?THANK U a lot

[@andrescrespo2514]

ilyass ourahou we assume it is massless and frictionless for simplicity’s sake, having the mass of the pulley will increase all the forces needed and friction will also, higher friction means higher inertia which means the pulley will not move and vice versa.

[@carultch]

If the pulley has weight (m*g), you also have to account for its rotational inertia (I) and the relation between rotational mechanics and linear mechanics. Unless you assume that the pulley's weight is concentrated at the axle, and has negligible mass at its edge. Rotational inertia is a property that depends on: mass shape distribution of mass and axis of rotation For a uniform circular disk of mass m and radius R, it is 1/2*m*R^2. There are formulas you look up for the rotational inertia of various shapes.

[@JettyBuilder]

Suggestions: 1. state that the tension in the rope is the same in all parts of the rope in a frictionless system 2. State that the global system has to be in equilibrium so in this example you have a 98N weight plus a 49N force on the end of the rope (being counteracted by an anchor of some type not shown) and these are in equilibrium with the 147N reaction at the top anchor

[@satyamkumar567]

Sir please tell, what book is best for mechanics ??

[@Turbohh]

nicely done, thank you.

[@MichelvanBiezen]

Glad you liked it. 🙂

[@kevinshaughnessy7225]

Thank you very much for sharing these videos. They're really helpful.

[@lookafterurlife]

Great . Thank you.

[@MichelvanBiezen]

You welcome!

[@vittoriopolverino1889]

Thank you, perfect explanation!

[@CristopherMenendez]

why wouldn't t5 be pulling on down t4?

[@MichelvanBiezen]

It does, but to be able to figure it out, you draw the free body diagrams as shown in the video.

[@matanlevi4457]

Thank you! from Israel

[@MichelvanBiezen]

Welcome to the channel!

[@matanlevi4457]

@@MichelvanBiezen thank you, may I ask if videos of specific subject are listed anywhere? like If I would like to learn linear algebra from scratch with your (awesome) channel?

[@MichelvanBiezen]

If you go to the home page you will find the playlist: HOW TO FIND PLAYLISTS ON THIS CHANNEL

[@MichelvanBiezen]

There are several playlists on linear algebra.

[@subasan4798]

Thank you so much 😊😊😊😊

[@benryckman3046]

Thank you so much

[@natashashepherd5690]

you are wrong in your final statement a force to move the 10 kg block must be great than 49 newtons in order to move it because 49 newtons of force the block is static

[@MichelvanBiezen]

If the block is moving at a constant speed, only 49 newtons are required. If the block is stationary you will need a momentary force greater than 49 newtons after which only 49 newtons are required

[@thakornjareonrat4450]

Thank you i can do homework now.

[@MichelvanBiezen]

Glad we could help!

[@PolkadOfficial]

the last fact was interesting:)

[@milanspasic1968]

Bravo

[@lookafterurlife]

Great

[@MichelvanBiezen]

Glad you found it helpful.

[@deref2857]

thank you for the explanation

[@MichelvanBiezen]

Glad it was helpful!

[@Dailymailnewz]

And many Amps will be 49 Newton if it can be converted? Any body?

[@carultch]

You would need to know the details of the motor and spool that are applying tension T1 to the system, if it is a motor in the first place. Otherwise, amps are not an applicable unit to this problem.

[@kyliechesner8933]

can u elaborate on why T1 = T3 ?

[@MichelvanBiezen]

Kylie, Since T1 and T3 are pulling on the same string, it must be the same, UNLESS the pulley has mass or has friction. Then one will be greater than the other because the extra tension is needed to overcome the friction or to overcome the moment of inertia.

[@AbuOmar1970h]

tank you helpful

[@johnnysparkleface3096]

So in a nutshell, this is a 2:1

[@MichelvanBiezen]

Yes, this pulley system has a 2:1 mechanical advantage.

[@yummyfoodvlog2238]

Fd

[@codingwithali2924]

ياريت تتعلم عربي وتشرحلنا ❤

[@ambikaraiswal4589]

so easy

[@lakshmisanthosh8429]

I guess the light there seems to be dim Please innovate good sources of light

[@MichelvanBiezen]

Yes, our older videos were not as good. We had to learn a lot about producing videos. The newer ones are much better.

[@victortsykunov]

How much t.ts. Kids never will understand poor little things eeeh