Two masses hanging from a pulley | Forces and Newton's laws of motion | Physics | Khan Academy

In this video David explains how to find the acceleration of two masses hanging from a pulley (using the easy method).
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[@clarisaleu-rodriguez5314]

Thank you for showing the easy way of how to do this

[@guzmanxp4906]

Do it the hard way man. Professor wants to see the hard way.

[@gaetanomanino1932]

did you finally find a video explaining the hard way?

[@BudderB0y2222]

Here's how u do it the hard way (m1=5kg, m2=10kg):
1. Find the weight of both masses using Fg=m*g; m1=49N, m2=98N
2. Determine which way the system accelerates (in the direction of the greater mass, so m2)
3. Use newtons second law, Fnet=ma, to find the net force for both. For example, Fnet=5kg*a for m1 and Fnet=10kg*a for m2
4. Find the second net force equation for both by subtracting Ftension from the greater weight (m2) and subtracting the smaller weight from Ftension (m1); Fnet=Ftens-49N (for m1), Fnet=98N-Ftens (for m2)
5. Set up a system of equations
Ftens-49N=5kg*a
+(98N-Ftens=10kg*a) -Ftens cancels out
49N=15kg*a
3.27m/s^2=a
6. Find net force by plugging back into one of your equations, like Fnet=5kg*a
Fnet=5kg*3.27m/s^2
Fnet=16.33N
7. Plug net force into an ftens equation and algebraically solve
Ftens-49N=16.33N
Ftens=65.33N

[@vanshsharma6809]

This has a solution too: ruclips.net/video/QKXeZFwFPS0/видео.html

[@rdl1ofri]

Great lesson. ty

[@audreybrien9495]

this was so helpful!!! thank you :)))))

[@visnjapetrovic5566]

in which program did you do this,it is nice for writing

[@HA-zg4fi]

this was so helpful thanks

[@naveedsyed2917]

Ngl i really like david's lectures but the "hard way" is much more easier and way less confusing than the easy way

[@zaynehasgills4104]

thank you

[@carloschuecos7014]

Thank you Sir, very helpful. What if the whole system is falling down?

[@sohamghaisas414]

Then all will be stationary relative to each other. Because gravitational acceleration is equal for all objects.

[@anmolaryan5840]

How will we calculate tension if a mass of 1 kg is more attached to 3 kg mass an we have to calculate tension between mass of 1 kg and 3 kg block

[@karp_4902]

If we have already designated that clockwise is +ve, then the final acceleration of the 5kg mass should still be +ve since we designated that direction to be +ve though?

[@zoeyoung7293]

this saved me from failing physics thank u my man

[@pats143]

thanks

[@sylviemuhimpundu9958]

Why is the acceleration on the 5kg box negative if the acceleration is going in the same direction as the motion?

[@matthewjames7513]

why did you assume that the reactions have no external forces??

[@vanshsharma6809]

Hello Khan Academy team, I just love your way of explanation and have your app too. But the only thing I wanted to ask was: Can we interchange the signs of acc? Like here you have given (-ve) for 5kg and (+ve) for 3kg, can we write that down as (+ve) for 5 kg and (-ve) for 3 kg, as the direction of acc is from 3kg mass side to 5kg mass side? Hoping for a quick response, thank you, and best of luck.

[@gracelytax653]

Not a member of the team at Khan Academy, but I can quite confident lying say that even if you do interchange the signs you will still end up with the same answer. Nevertheless be careful when writing your final answer!

[@fritzclark082]

How did you get 36.75N..I just needed to see it clearly please without calculator. .thanks

[@mehranahrari3746]

-2.45m/s^2 * 5kg then 5kg*9.8m/s^2 and move it over to the other side to leave T on its own gives you 36.75N

[@mojoman7141]

5 - 3 = 2 *9,81 / 8 = 2,45m/s²

[@sarahbaldwin3162]

Another way of thinking as internal forces as negative is the force is heading increasingly inward therefore negatively. While external force heads ever outward making its movement away from the system eterbal positive.

[@trentoncook8969]

4:56 how did you get 2.45M/s (squared)? I'm brand new to these types of problems and ready to kill myself over them.

[@trentoncook8969]

Nevermind, F = MA *facepalm*

[@thirdly5509]

What if the mass of the two objects are the same?

[@Fenrick]

This kind of calculation would not work because this assumes that the rope is weightless (which it is obviously not). In such a calculation, whichever side has more rope would pull the other one. But that's a lot more advanced than this video.

[@cameronsantiago3155]

The tensions will not equal each other throughout the string???

[@drofeng]

Yes. One way to think about this is to take free body of part of the string (e.g. between the pulley and 5kg box). Doing a force balance, FT1 - FT2 = m*a. If the string is treated as massless, FT1 = FT2

[@jennyhancock3474]

Could you please explain how you got 36.75? I've typed it into the calculator so many different ways and still got different answers every time :(

[@annah5299]

You would multipy the sides by the 5kg and then that would get rid of the fraction. then on the left side you would have -12.25 = T - 49 and then get rid of the 49, you will have T = 36.75

[@francisdelrosario9041]

@@annah5299 thank you

[@mohammedghothaya5119]

billion thanks

[@ahmadbelhaj1756]

what if there is friction on the pulley

[@adityasapra3419]

Ahmad Belhaj as he said we avoid the internal forces so friction will be negligible

[@josicane1624]

So what about the hard way

[@bleached_coral]

set equations F=ma for both objects, with T and a (same acceleration for both objects) in them. it will usually look something like this: T - m1 * g = m1 *a; m2 * g - T = m2 * a. solve the simultaneous equations then

[@ricaramirez5411]

I did not understand the solution for tension. I got different answers everytime I solve it. Send help pls :(

[@ricaramirez5411]

oops i got -36.75

[@ricaramirez5411]

oh nevermind i got it already thanks!

[@jackfiddleton]

relax god damn

[@burlofreak]

Your "easy" way is the exact same as your "hard" way!

[@aswfasho1674]

I’m confused

[@nicolas.leger3]

So weird 😅

[@sneakyswiper3831]

the acceleration should stay positive not negetive

[@dobeeeeval]

You go from one second saying the tension caused by the 3 kg block doesn't factor into external forces calculation, and then immediately follow that up by calculating tension caused by the 3 kg block and subtracting it from the tension caused by the 5 kg block to calculate external forces... Obviously I just don't understand what you actually mean, but you seem to contradict yourself. Maybe provide some clarification.

[@maryclairebarton4911]

yo this pulley massless or nah

[@BudderB0y2222]

yep, or else you'd have to include it in your net force

[@KillEveryWokeLiberal]

Sal is better

[@salwarafi9812]

DAVID IS BETTER

[@crztrod]

fuk u

[@kritikasharma5436]

@@crztrod i knew this would be there

[@weeklyhunt6379]

Wasteeeeeeeeeeeeeeeeeeeeeeeeeeeedeedeeeeeeee