Physics Lesson: Pulley Systems Part 2 Dynamics for High School

[wcwc89]

I was in his class 5 years ago, and going in I did very poorly in gr 11 physics with another teacher, I was fortunate to have him as my teacher in gr 12 and finally understood it all with his explanations. I did extremely well in his class too

Great teacher and might know a thing or two about playing dodgeball ;)

[caseykc18]

BRILLIANT.... Honors physics final tomorrow. This helped me a great deal.

[TheMrShah1]

one question what kind of a fag hates a free physic video?

[Dragonxsolo]

SO BEAUTIFUL

[Klegend1000]

Can you make a video on movable pulleys? cuz i saw a picture of a movable pulley on our book but i didnt understand it. And I am not highschool yet so can you make it a little more simple? cuz i dont know anything about the equations.

[warnford]

wonderful

exxaxctly as my maths teacher Fatty Barrett of Beaminster School taught me in 1968.

[SkyEZ653]

Thanx for the great video. It really helped me out.

I just wanted to know. How would you resolve about the 10kg. The tension in the string would be the same but what about the acceleration??

[daemonpacman]

I don't know how apparent this is to you, but you are doing a great service to the world by making these videos. There are many students who are struggling in their physics and math classes and just want to give up. But then they see these videos which present a very clear explanation of the concepts they are struggling with, and press on. I wonder how many college and high school dropouts you have prevented...I guarantee it isn't zero.

[mrtsunny]

okay, this makes a lot of sense now. thanks!

[RescuePits1]

I like how you simplify it (4:11), it makes it easier to grasp the concept.

[jeysidalumpines]

why does the y direction in 5kg is equal to zero? :)

[1917warfare]

What would the tension on the rope be is the surface was frictionless?

[PhysicsEH]

@1917warfare It should be about 3.5N Hope this helps

[rip111buddy]

@1917warfare @PhysicsEH 10g - 0 = 15a

a=10g/15

T-0=5a

T = 5*(10g/15)

=> T= 33.33333333333....N

[BkSenko24]

Looks like someone is going to pass AP physics....

Thank you I almost cried this was so beautiful

[PhysicsEH]

@BkSenko24 Thanks, that comment is a keeper. glad we could help.

[BikutaHo]

thanks for the video

[PhysicsEH]

@BikutaHo Thanks. Glad to help.

[HMShafqat]

really helpful stuff.



[PhysicsEH]

@HMShafqat Thanks. Glad to help.

[browndolLxoxo]

this was SOO helpful thanks!!! :)

[PhysicsEH]

@browndolLxoxo Thanks for the comment. Glad we could help.

[7hrjpidocil]

I'm glad that this video along with your others are concise and informative! Keep up the good work.

[PhysicsEH]

@7hrjpidocil Thanks, we love to hear that.

[SuperSkying]

well, what if the object on the table have greater mass? what would the system of the motion be? which way will the acceleration directed?

[PhysicsEH]

@SuperSkying Hi Sorry I missed your question. The only way the object can accel is toward the 10kg because gravity is pulling on it and there is no other force that is pulling the other way. The gravity of the other object has no component in a direction that the object can accel. Friction can slow it down but can not pull the object along the table. If the mass on the table was larger it would accel less and at some point the Ff caused by the mass will cancel the gravity and it would not move.

[newage4energy]

Its gonna take me awhile to watch all these interesting videos. Thanks for taking the time.

[PhysicsEH]

@newage4energy Thanks for the great comment. Glad we could help and I hope you find them all good.

[rip111buddy]

hey...a question.....do you take g=9.8m/s[square] or you just round it off to 10m/s[square]?

anyways.....YOU ROCK !

thanks got the sooperb videos..!!

[mcaruana1]

@rip111buddy We round off to 10 m/s^2 when we do pulley problems because it keeps it simple. We do this a lot for "g" for gravity also. Glad you like the videos. Thanks for letting us know.

[lanzxy]

Nice video. happy to learn physics with your videos. Keep it up. :)

[PhysicsEH]

@lanzxy Hi. Thanks for the kind comment. Glad you enjoy them.

[yellopanda]

thank you ! you were great in the hangover, Zach !

[PhysicsEH]

@yellopanda Thanks. I'll let him know.

[Orijahnalsound]

holy shit yo. i got an exam this mornin been up all nite tryna remember how to do these. appreciate yu puttin these up. dont stop

[PhysicsEH]

@Orijahnalsound Hi. Thanks for the kind comment. I hope your exam went well.

[ford26993]

Sorry, but I don't understand the part where you put Ff = mu*Fn . . .

Because isn't mu the friction force? So isn't that like saying friction force = friction force * the normal force... Not quite understanding that bit. If anyone could help to better explain it, that would be great :D

[PhysicsEH]

@ford26993 Hi. mu is not friction but the "coefficient" of friction. That means it is the number you multiply Fn by to calculate the friction. Ff = mu*Fn I hope this helps.

[ford26993]

@PhysicsEH ok thanks heaps! I don't remember ever being taught that or doing questions with a coefficient but thanks. I'm from Australia and I did my exam today. Big thanks to you guys for making things a bit clearer for me :)

[TricolourPictures]

Thanks very much! This will help with my final tomorrow. Going for 90!

[PhysicsEH]

@TricolourPictures Thanks for letting us know they helped. Good luck on your exam!

[MiturBinlsderty]

man you are the boss! love how i can easily follow the work. love the clean writing and clean chalkboard. love it. love it. thanks.

[PhysicsEH]

@MiturBinlsderty Thanks for all the compliments. We are glad you found them helpful.

[tonydeadlock]

what is the mu thing

[PhysicsEH]

@tonydeadlock That is the coefficient of friction, which is a number that lets you calculate the friction on an object. It is usually given in the question. Hope this helps.

[ahlawy1993]

Clear, short and helpful.. Thank you very much

[PhysicsEH]

@ahlawy1993 Glad to help.

[Climate911]

Hold up. I got the acceleration to be negative. IDK if I did it right but this is what I did:

a = 10(-9.81)-.2(5)(-9.81)/15

a = -98.1+9.81/15

a = -5.88m/s^2

This is also confusing for Ft.

Ft = 10(-9.81)-10(-5.88)

Ft= -39.3N

Can you give a clear concept on negative signs.

[PhysicsEH]

@Climate911 The answer will be in several replies.

The best way to do this is to make the direction of acceleration = positive (to the right and down is then +) so the force of gravity on the 10 is down and thus + and the force of friction is – because it is to the left.

NOTE: the 9.8 you use to calculate the gravity on each object is always +. The 9.8 is just used to find the magnitude of the gravity.

[PhysicsEH]

@Climate911 Part 2

You then make the force of gravity + or – depending on which way the force is acting, so the force on the 10 kg is 10 x 9.8 = 98 and it will be + because the force is down on the 10kg and we made that direction +.

The force of friction is 0.2 x 5 x 9.8 = 9.8 and it is negative because it is to the left on the box and we made to the right +.

[PhysicsEH]

@PhysicsEH Part 3 In your answer you put in – for the 9.8. Don’t. So you get a = 10(9.8) – 0.2(5)(9.8)/15 a = 5.88 For Ft you get your equation from Fnet = ma Fg (down is +) – Ft (up is -) = ma Fg – Ft = ma 10(9.8) – Ft = 10(5.88) Ft = 10(9.9) – 10(5.88) Ft = 39.3N If you get negative answers it usually indicates you made an error. I hope this helps.

[PhysicsEH]

@Climate911 Hi. You may have to go to the video to see all 3 parts of my reply. Best of luck.

[Climate911]

@PhysicsEH w00t FINALLY got the concept. Thanks.

[PhysicsEH]

@Climate911 Great. Glad to help.

[crimsonknight252]

Dude you are awesome!