Physics 4 Newton's Laws of Motion (18 of 20) Statics: Example 1

[@gjdgc3771]

Man, to all who benefited to this. dont forget to click the like.. You're a saviour sir!!

[@annalietrodriguez4787]

I just want to say thank you your videos are so good and I can understand easily .

[@historyisthebest5831]

I was not able to deal with this kind of trigonometric physics problems(I haven't formally learn trigonometry in school yet), but watching your vids gave me the ability to cope with these problems.

[@mickclark5030]

What a difference somebody knows what he is taking about and not from a note book. He took me back to 1964 and what I tried to understand was made so clear that at 71 years I fully I stood. Thank you very much

[@MichelvanBiezen]

Way to go! It is exciting to finally understand these concepts. I get the same kick out of finally understanding concepts I didn't understand many years ago in the classroom.

[@rawrz123]

Man this dude a legend

[@MyMusicCreatesMe1]

SOOO HELPFUL!! I can't thank you enough Mr. Michel van Biezen! Thanks for all the time and effort to put these videos.

[@Jcakiiiii]

Thank you!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! i learned so much with all your videos! :D

[@tona017]

You made this example so easy to understand. Thank you.

[@heronkulukul4156]

This is helpful, thank you. But how can we find the size and the direction of the resultant tension? (For the same problem)Thank you

[@MichelvanBiezen]

If you pick the point where all three ropes meet, the sum of the forces in all directions will add up to zero. (Therefore we don't find the "resultant" tension or force since it will always be zero at any point under consideration, since it is a static situation).

[@MichelvanBiezen]

Albert Geller:The video is in the following playlist:PHYSICS 4 NEWTON'S LAWS OF MOTION

[@gjdgc3771]

is this can be done with vector analysis methid? because its the same problem but (only with different values) in my vector analysis subject..

[@abirdas3955]

We also can slove this types of equilibrium problems by Lami's theorem. I love to use Lami's law to slove equilibrium problems

[@MichelvanBiezen]

Yes, that is a perfect valid method to work out this problem

[@historyisthebest5831]

I seem to understand this question but I want to make sure. This question is also asked somewhere else. Why do we neglect the components of Fx forces and only calculate components of Fy forces?

[@MichelvanBiezen]

The forces in the x-direction are not being ignored. They are being used to help find the solution.

[@historyisthebest5831]

@@MichelvanBiezen Thanks! But why don't we count it when calculating the net force?

[@carultch]

@@historyisthebest5831 We do. We just count it in a different direction, because we develop a force balance equation in both directions separately. In the horizontal direction, only the horizontal components of the two diagonal strings apply a force. In the vertical direction, the vertical components of the two diagonal strings apply a force, as well as the weight of the supported object. In the direction perpendicular to the whiteboard, there are no forces, so we don't have a calculation in that direction to even think about. Since this is a statics problem, by definition forces add up to zero, so the net force is zero as well.

[@historyisthebest5831]

@@carultch Thank you.

[@ALAL-jm1lr]

plz sir, does this mean the smaller the angle the bigger tension on ropes? and who exactly contribute to these wasted forces? the wall?(cuz mg is mg....no change on weight...)

[@MichelvanBiezen]

Indeed as the angle gets smaller, the tension increases and goes to infinity as the angle goes to zero. (I wouldn't call them wasted forces)

[@ALAL-jm1lr]

@@MichelvanBiezen thank u for ur reply, actually i try to explain those forces as being needed to stop the mass from swing to left or right ,so it just hang there in that exact middle lower location ....physics is insane!

[@petermaroney3625]

Great video. How can I solve one of these problems if one of the angles are not given?

[@MichelvanBiezen]

You need the angles, otherwise you cannot solve the problem.

[@paulhunt9533]

thank u sir i understand easily

[@erc333]

... Why don't mg and Ft3 cancel out if the system at rest? Which would mean: F net y = Ft1y+Ft2y+Ft3-mg ...therefore, F net y = Ft1y+Ft2y ...?

[@MichelvanBiezen]

Whenever you add forces, you must pick a single point and then add all the forces. Picking the point where the 3 strings come together you can say: Sum Fy = 0 = + T1y + T2y - T3

[@erc333]

That makes sense but, if you're referring to the forces on that certain point where the three tensions "connect" then wouldn't we have to use the mass of that certain point and not the 2kg mass?

[@ElegantStrife]

The mass of the line is negligible in the calculation; in reality it makes a very small difference so when doing these calculations they are ignored for simplicity.

[@zakariafatima9105]

How do i solve equilibrium of forces on incline planes?

[@MichelvanBiezen]

We have many example videos on the inclined plane on this channel. You can find them all from the home page.

[@boredasever6645]

you are truly a hero

[@parkerd2154]

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[@jamesbanda3944]

I thank you so much for your videos, they have been helpful to me

[@vinitabhadoriya153]

Awesome thank you so much this cleared out my doubts

[@graceb2404]

Once again, thanks!

[@albertgeller]

This video inst indexed in your physics mechanics playlist, so please put it on there. Thanks for the videos, it helps me a lot =), so keep doing this work.

[@thailandfutsal5508]

If I want the system to be in equilibrium and I calculate net force to be 0 but how do I know that moment EQUAL 0

[@thailandfutsal5508]

Hi

[@thailandfutsal5508]

Please describe

[@MichelvanBiezen]

If the net moment is not zero, the system would start rotating. That fact that it is not rotating indicates that the moment about any point MUST be zero.

[@carultch]

@@thailandfutsal5508 In this particular problem, we aren't interested in the net moment on any object, because no object has significant size other than the string segments. Because the strings have no significant mass of their own, they are two-force members, which means the only force in the strings are the equal and opposite tensions that add up to zero on each string. Because the tension forces on both sides of the string have lines of action that are both colinear with each other, the moment of both of them adds up to zero. If the strings had a mass of significance, we'd have a catenary curve problem to solve, when considering how a hanging cord supports its own uniformly distributed weight.

[@drdin3442]

Guys you can use this formula which I made. T1=(mgcos(theta2))/sin(theta1+theta2), T2=(mgcos(theta1))/sin(theta1+theta2)

[@erikm5753]

Would, or how would, the tensions change if the load was hanging from a pulley on the T1/T2 line? Assume the two angles are different because they are hanging from different heights.

[@MichelvanBiezen]

Erik, There is a video where the angles are different. If the masses don't move it doesn't matter if there is a pulley.

[@erikm5753]

Michel van Biezen Thanks. I'll look at the next video, and comment there.

[@briankylegarsuta1190]

What if both sides are 55 degrees, is the procedure still be the same?

[@MichelvanBiezen]

Yes, it doesn't matter what the angles are, we use the same procedure.

[@briankylegarsuta1190]

Okiee,Thankyou!

[@Asha-ll1sp]

Amazing video! Thanks

[@amarynthia2990]

what of there are 2 masses/blocks?

[@MichelvanBiezen]

That would make the problem more interesting (and a little harder). We should probably make some videos like that.

[@holyromanempire7555]

This was so helpful!!!!!! Thank you!!!

[@maiaxxx8318]

THANK YOU SO MUCH

[@PK-by4ii]

i have another ideas professor can we cal sigma f=zero . t=mg sin45

[@jaicahermogeno4021]

i was so frustrated coz i feel like i was the only one in our classroom who doesn't understand this and then I saw this just wow

[@MichelvanBiezen]

Seldom you will be the only one in the classroom that doesn't understand something, but everyone seems to be afraid to ask the question. (I used to think the same thing when I was a student).

[@rahultiwari9003]

sir when a body slides over another body can I conclude that the body is accelerating

[@MichelvanBiezen]

+Rahul Tiwari No. For a body to be accelerating it must experience a net force. (F = ma) Newton's second law.

[@amandabekka6636]

Amazing!

[@dndeveloper]

What we doing in chain hanging problems Sir Please help me

[@carultch]

A connection cord of significant mass (whether it be a chain, a cable, or a rope) will make the problem much more interesting, because the cord will no longer take a straight line path when supporting a diagonal tension. Instead, it will follow a catenary curve, and you will need the background of differential equations to derive this. There are catenary curve equations that are derived for you, to determine the sag curve and the direction of tension on both ends, as it relates to the initial length of the cord and the weight of the cord. The tension will also not be equal and opposite, like it is in a cord of negligible mass.

[@jakurdav]

or we could just find components of F, mgsinTheta and mgcosTheta which are equal to T1 and T2, because they are opposite :)

[@hamidirn1686]

What if we don't have the mass but everything else is the same? The question is the same

[@carultch]

You have to be given something that applies a load to the system. You might be given this problem in reverse, where it tells you the tension in one of the strings, and you have to solve for the supporting mass. In any case, you would assign a placeholder for the mass of the hanging object, set up the problem as you normally do, and then solve for the unknown. You might also be given a human force, instead of a hanging mass. The solution procedure is still approximately the same.

[@darrelltieu5934]

ty

[@pranavamali05]

Thanku

[@MichelvanBiezen]

you are welcome

[@arbysandtehchief5494]

Thank you!

[@jahanzaibkhan2587]

If T3 is not equal to T1 and T2 then how to calculate T3?

[@MichelvanBiezen]

T3 is the equal to the weight hanging from it.

[@harjitkaur1601]

Fnet - ma= 0 Considering just the object and the tension in the string, the forces acting in the y direction must be 0 since sumFy = 0. Thus T3=mg

[@ajilbabu13]

@@harjitkaur1601 thank you

[@ybrikjosh]

I got 11.53 newtons?

[@MichelvanBiezen]

For which calculation?

[@Jayc700]

thanks!!!!!!!

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[@sravanmudiraj677]

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