Physics - Mechanics: Applications of Newton's Second Law (14 of 20) inclined plane

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In this video I will show you how to calculate the acceleration of 2 masses around a pulley attached to a wedge.

[@TeenGohan9798]

at video 1 i didnt know anything , now i come here and solve it without even watching the video 1st ,YOU ARE A GOD

[@MichelvanBiezen]

No, no, just an ordinary person, that is glad that we can help.

[@signalcloud9202]

Saved my life prof. Thanks!

[@MichelvanBiezen]

Glad it helped!

[@khylleiandalida3651]

Thank you, sir Michel, you saved me again

[@MichelvanBiezen]

Happy to help

[@venggkadesh2947]

hi,why must we minus the component of the weight which is parallel to the surface when you already minus the frictional force? (4:00)

[@MichelvanBiezen]

They are both independent forces and both act in the opposite direction of the acceleration. (They oppose the acceleration).

[@venggkadesh2947]

thank you! great videos :)

[@mikemai8568]

Your video is great and easy to understand. Thanks

[@deryaulu8902]

this chanel and the video so good , ı love the teacher ‘s explain , thank you very much

[@zeyneptufan5789]

Hi, sir. In last step why didn't we write F friction? I mean the equation m2.g-T=m2.a

[@sylvsc.9434]

This is really helpful! Thank you so much for making physics so much easier :) !!

[@عمروابراهيم-ط3خ]

Hello Michel, while getting (T) why dont we use (Fnet=(m,total)*a) like the first?

[@MichelvanBiezen]

In this case you are not finding the acceleration of the WHOLE system, but the tension acting on a single component of the system. That is why you use a free body diagram (adding all the forces acting on a single component) and then solving for T

[@davidcruz3400]

Great Video!

[@pascalfroehlich8013]

in order to get acceleration you defined a=F(net) divided by m(total)...this F(net) ended up being 81 Newtons (which I thought would make a nice tension force) then you define F(net) as m2g - T which if T ends up being 40N and m2g is 10 times 9.81=98.1N would only equal 57.9N 98.1N-40N...why do we have 2 F(net) with different values?

[@MichelvanBiezen]

We start with the same basic equation, (F = ma), but they are not the same Fnet. The first Fnet is the net force acting on the system. The second Fnet only applies to force acting on the m2 mass.

[@pascalfroehlich8013]

@@MichelvanBiezen So... kind of like Fnet1 and Fnet2... Gotcha... Thanks boss:)

[@jerome3745]

Sir. In Which référentiel do you calculate a when you solve Fnet=Mt*a ? You add m2g and m1gcosalfa+Rt, however these forces are not in the same référentiel. Is it possible ? (Sorry for my very bas english)

[@jezzanabong9428]

please recheck the value of the kinetic friction. is it really 3.7? 4(9.8)cos20 (.1)

[@jezzanabong9428]

should i set my calcu in degree mode?

[@MichelvanBiezen]

Yes, your calculator should be in degree mode (since the angles are expressed in degrees)

[@roberttokona382]

just made my day.. thank you very much..

[@iqrajamal4361]

wonderful MR BIEZEN U r GREAT 👌👌😃😃

[@kingonion2102]

Why can't we solve for the acceleration in each block and then combine the equations to determine the tension? When I try to do that I get a tension that is smaller than 40N

[@MichelvanBiezen]

You can solve it that way, but it will be more difficult and more complicated. You end up with multiple equations and multiple unknowns. Much easier to solve it as shown in the video.

[@kingonion2102]

@@MichelvanBiezen Hey Michel, thanks for the quick response. I also figured out why my tension was wrong (my friction was positive instead of negative). I love all the work you do, keep it up!

[@MichelvanBiezen]

Thank you. We plan to.

[@striker800]

Shouldn't cos be sin because it is opposite from the angle theta?

[@MichelvanBiezen]

The sin and the cos in the vido are correct.

[@ptyptypty3]

Hi Michel, if M1 were a Wheel and not a block, could we still use the same equation for Friction force without using the Torque Equation?

[@MichelvanBiezen]

Hi Philip. If the wheel slides up the hill, yes. But if the wheel rolls up the hill, no. We have to take into account the moment of inertia, and how / where the wheel is connected to the string.

[@ptyptypty3]

@@MichelvanBiezen Thanks Michel. I'll use the usual Friction Force = (1/2) ma For a Wheel. then redo the calculation.. :) thank you!

[@factstechandtricks5769]

that mean tension in the string is independent to friction ?

[@zeyneptufan5789]

I also didn't get this thing clearly

[@rustamshahverdiyev4164]

Professor i have one more question. As i understood aiding and opposing acceleration depends on mass of object, i mean for example if m1 will be bigger than m2 then m1 will be aiding and m2 opposing. Are my arguments correct?

[@MichelvanBiezen]

In this example, m2g is the "aiding force" and m1g sin(theta) and m1g cos(theta) u are the opposing forces.

[@Sageboy13]

Shouldn't cos be sin because it's going down in the Y direction?

[@MichelvanBiezen]

Cos is correct. Don't associate "down" with "sin" automatically. Use the definition of sin an cos. Cos = adjacent side / hypotenuse

[@KenzoKento]

hi michel, why do we not include tension for fnet? do they get cancelled out by each other?

[@MichelvanBiezen]

+:3D The tension is internal to the system (the force between two bodies that make up the system). Only forces external to the system need to be considered. (Unless you draw a free body diagram of a single body of the system, then you must include the tension).

[@elenirigopoulos4535]

Very easy to follow, thank you so much!

[@yamnagirl6837]

Brilliant sir. Thanks.

[@JimTheune]

Thanks! You make good sense of this topic.

[@younique9710]

Sir, I have a question on this problem. At 3:51, why is the sign of the fractional force is minus, rather than plus. Because the frictional force is the opposite direction from the m1g*sin20, must the direction of the frictional force be the same direction with the m2g? like [m2g + (the friction) - m1g*sin20] / (m1 +m2)?

[@MichelvanBiezen]

We use the sign convention: 1) if the force aids the acceleration, the sign of the force is positive. 2) if the force opposes the acceleration , the sign of the force is negative

[@koketso_nk]

Sir, do you have a problem where you use both static and kinetic friction on incline?

[@MichelvanBiezen]

There are two playlists on friction: PHYSICS 4.6 FRICTION PHYSICS 4.7 FRICTION & FORCES AT ANGLES kzbin.infoplaylists?view=50&shelf_id=4&sort=dd

[@koketso_nk]

Thank you!

[@zameerahmad8502]

it was helpfull thank you

[@sanjaykoonoolal523]

how would you find velocity from this?

[@MichelvanBiezen]

If the acceleration is constant you can use the equations of kinematics. V = Vo + at etc.

[@sanjaykoonoolal523]

thank you

[@tanleminh4281]

very well, thank you so much !!

[@MichelvanBiezen]

You are welcome!

[@Zombeizz]

what if we don't know μ?

[@SSSJ0014]

Physics made easy

[@nobledzas3690]

thanx alot

[@sugithkandiah457]

faster than my teachers method

[@sbs6259]

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