Physics 4.6 Friction (12 of 14) Direction of the Friction Force: Ex. 2

In this video I will find a) F(fr)=? on m1, b) F(fr)=? on m2, c) T=?, and d) a=? of m1 of m2 on top of m1 with m1 and m2 attached to a pulley which is attached to a wall with a rope and m1 is pulled in the opposite direction with force F.

Next video in this series can be found at:

[pipertripp]

Quick question for you.... I notice that you didn't determine the Tension. I'm guessing that since you're treating both blocks as the system, rather than considering them individually, that you can ignore the tension because it's an internal force, correct?

I treated them individually so I actually had to calculate the tension before I could get at the acceleration. I used a = Fn/m for each block and knowing that they must accelerate at the same rate (since they're connected by the cable) I could equate Fn/m1 = Fn/m2 and that allowed me to solve for the tension. Once I had the tension (massless pulley to tension equal on both sides of it) I could just plug that value back into the acceleration equation for one of the blocks and get the acceleration from that.

The tension magnitude I got was 33.83N... I got the same value for acceleration as you so I figure that tension is correct.

Cool problem. I'm surprised it hasn't gotten more views.

[eswarawakasimonsahith]

That was a very great explanation Maichael sir, because very few lectures of physics will explain as clear as you sir.

[DevonRavihansa]

Sir, I have calculate
▪The Tension of m2,
(accelerate upward = leftward '+ve')
Fn=m•a
T-Ffr3=m2•a
T-m2•g•u2=m2•a
T=m2•g•u2+m2•a
then I got, T=33.72N as results
and
▪The Tension of m1,
(accelerate downward = rightward '-ve')
Fn=m•a
(T+Ffr1+Ffr2)-F=m1•(-a)
T=m1•(-a)+F-Ff1-Ff2

then I got, T=34N as results.
Those are approximately equal but slightly differs.
Am I correct?? assumptions and results.

[orangeyouafunnyone]

I probably misheard but I think at 2:27 you said "All divided my u1" ?

[eswarawakasimonsahith]

A doubt sir, :How to calculate the tension sir?

[susmitasunar]

Thank you so so much sir...you helped me a lot..🥺❣️❣️

[aminhadi]

i am assuming he didnt include the tension forces bcs tension 1 is equal to tension 2 bcs the pulley and the string are mass-less and friction-less but they have opposite direction

[ahmedal-ebrashy]

My value for tension was tension equal mass of box two times acceleration of the system + friction in the opposite direction of tension cause by box 2 on box 1 which equaks 33.819 N please tell me my answer is correct. And thank so much.

[kniflapunxFucknazzi]

can i write that tension for object 1: T1 = ( F - Ffr1 - Ffr2 ) - m1a ? Is tension 1 equal to tension 2?

[tunajoseph]

why we added F3 friction force in the opposite direction when we found acceleration. Isn't it in the same direction with F?

[ayoolu]

If force, F is applied to both m1 and m2 simultaneously, will the frictions fr2 and fr3 act in the same direction?

[gilbert]

Could you do a video of this exact question calculating the tention?

[M_.a_.l_.i_.k]

Tension force screams in the corner 🥲😂

[spsi]

If there is a pulling force on the top block instead would you use the same formula seen as there are the same frictional forces?

[pacop.o.]

If there was a mass of 1kg over 2kg, with a pulley with a mass of 4kg with coefficient of friction of 0.05, how will it affect in this system ?

[sthelilesihle]

Sir, would it be wrong if I set up equations for the two blocks separately and then add them to find the acceleration?

[joyli]

Can you explain how to get T base on the acceleration?😥😥😥

[annaxzl]

Isn’t tension also an opposing force for m1？should it be included while calculating acceleration？

[abdullahrazzaq]

Thank you very much for the video lesson. i have a question relating to the normal forces. When the count the magnitude of the normal forces they should cancel out the magnitude of the gravitational force no? in this case m1g+m2g+m2g does not equal m1g+m2g

[Rohitkumar-yojs]

sir why you did not calculate the value of tension there??