Physics 4 Newton's Laws of Motion (20 of 20) Statics: Example 3

[@MichelvanBiezen]

You made a very good observation. Yes it is not possible to pull a cable into a straight horizontal shape. That is why power cables have a certain amount of sag. But you have to balance it out. If you make them sag too much, the cables are much longer and that costs more and also the weight increases which will also increase the tension in the cable. It is an interesting problem to figure out what the correct amount of sag is.......

[@wasafitvonline9248]

Michel van Biezen sir.. I just find the solved problems on this topic but in these cases 1. exploded bodies 2.rocate 3. water from the water pump to the wall 4. accelerating sand into lift.. and such stuffs ... please can I get help sir

[@historyisthebest5831]

Sometimes we assume that the ropes do not break or sag when we're simplifying a problem to an ideal state.

[@bulutistaken]

Your teaching is legendary! Thank you professor, this video was the last before my physics 101 exam tomorrow. I believe I'll pass it with a good grade thanks to you.

[@MichelvanBiezen]

All the best on your exam!

[@wardaavery4401]

Omg I'm so happy right now, after four years I understand this kind of problems !! thanks a lot professor. Great work! Greetings from Belgium

[@MichelvanBiezen]

It is a small world. I grew up in Belgium (a long time ago). We zijn blij dat we konden helpen.

[@wardaavery4401]

Ja inderdaad! Haha ohh zoo leuk. En bedankt. Ik apprecieer het :)

[@chrisbrice9332]

Another great playlist, sir. 10 down, 130 to go😀

[@MichelvanBiezen]

So close!

[@sarasaidova2308]

So easy and understandable explanation I like it so much

[@MichelvanBiezen]

Thanks for liking

[@conservativedeveloper7289]

Well, i'm pretty much an expert on newton's laws of motion now. Thanks!!

[@jarryingnaut]

Michel van Biezen, You are just amazing!

[@fofoca1731]

I am very grateful I found your channel. Thanks!

[@ivanthegreat1004]

Thank you for your videos.. i was struggling in class, and now i feel more confident over this material

[@michaelle1897]

I wish you were my professor.

[@PrinceKr-x6k]

Thank u so much professor sir my all doubt clear and also being a good concept of mine about NLM ❤❤❤

[@MichelvanBiezen]

Glad you found the videos helpful.

[@PrinceKr-x6k]

@@MichelvanBiezen sir this side PRINCE and I watch daily your videos .. do you have remember I promise to you that I will watch your all videos under 1 year or more but I will definitely watch all videos ...😊

[@anoobgamer1459]

thanks man tmr is my test and that is really helpful :D

[@Zarni1994]

Thanks professor. Your lectures are amazing

[@Dharsha-p4h]

Hi sir We can get the value of T1 by solving the both equations So ,T1=mg/tan45

[@iveenpatrick6141]

I have learnt something good in this videos

[@samvidpundir2922]

Thanks, now i think i have gained some mastery over newton's law of motion

[@sophornsomoun1303]

Thank you very much for your good videos................that's really nice

[@SSSJ0014]

Thanks professor you been teaching me 👍🏼

[@benbarris3648]

You help me so much thanks doc!!!!

[@jemeleepagay9032]

You're my savior omg. I got it, thanks to you!!!

[@nabilahfarhanah3454]

why in x-component,-mg,the weight is in negative direction? it should be in positive direction.is it right?because the gravity is in positive direction.

[@nabilahfarhanah3454]

for this problem,it's okay.i already got it.btw thanks for the video.it really helpful.

[@mandeepchoutapelly2939]

Are their any videos on Friction too??thnk u for these...

[@MichelvanBiezen]

We are working on a playlist now just about friction, but this playlist has a number of examples with friction already. Take a look: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS kzbin.info/aero/PLX2gX-ftPVXV0Hc1CGnYKaPq19P3RX0Lx

[@yashchaudhary2787]

thanks!! for such an amazing playlist.

[@gloriacarvajal5021]

Thank you so much professor. Suscribed!!!

[@user-tm1ix7xi1n]

If we are separating T2 into sin and cosine, then why didn't we do the same thing for T1?

[@MichelvanBiezen]

T1 only has an x-component.

[@shahenazshaikh7795]

kshitiz rimal T1 is already on the x axis so no need for breaking into its components

[@MuhammedIbnSuhaylIbnAbdAlAzeez]

kshitiz rimal Because there was no angle for T1. Therefore there is no components

[@albertopoli8896]

But T3 isn’t upward? Or I’m doing confusion?

[@MichelvanBiezen]

T3 acts downward and that is why there is a negative sign in front of mg in the equation.

[@Junyuancreme]

Thank you sir!

[@thailandfutsal5508]

But when I calculate how do I khnow that force will make the moment be 0

[@MichelvanBiezen]

The net torque is always zero for any static system

[@TeenGohan9798]

how will i know that im not supposed to divide T1 to it's components?

[@MichelvanBiezen]

Vectors that are pointing in the x or y direction don't need to be de-composed into their components.

[@jadespencergonzales5775]

Hello there, how to solve the same scenario but the angle is missing

[@MichelvanBiezen]

Is T1 given?

[@jadespencergonzales5775]

I'd like to send to you the question can you help me. The 200 kg crate is suspended using the rope triangle A and Triangle C. Each rope can withstand a maximum force of 10 KN before it breaks. If AB remain horizontal, determine the smallest angle theta to which the crate can be suspended before one of the ropes breaks. Thanks God bless

[@MichelvanBiezen]

Set T1 equal to 10 kN and then solve for the angle. (For small angles, T1 will have the greatest tension.)

[@jadespencergonzales5775]

Can I request sir, I am a Filipino and glad to have your videos such a great help. Can you make a video out of that equation which I have sent. God bless you sir and more power

[@MichelvanBiezen]

It is an interesting problem, but it takes a while to get this through our production run. There are many videos already in the "pipeline" to be produced.

[@arnab_d9]

thank u so much

[@RM-ek4bg]

why is T3 downward?

[@MichelvanBiezen]

It depends on the perspective. Relative to the point where all three strings come together, T3 is downward.

[@RM-ek4bg]

sorry, I am still confused. I have a few questions: 1. Does that mean T3 is basically the weight force? I always thought that tension acts oppsotie to the force acting on the rope. 2. Is it correct to say that the tension T3 is the total vertical component and the tension then distributes to T2 hence T3 = T2sin(theta) = weight? (because we can assume that the rope is weightless and that the tension is same through out?

[@rajradhaiyan8196]

tq soooooooooooooooo much

[@joshuahor8528]

Thanks a lot! But it seems there might be some easier ways to solve it..........

[@thailandfutsal5508]

But how is khow that torque equal to 0 too

[@MichelvanBiezen]

If the system is static (not moving) then the sum of all forces in the x-direction = 0 , the sum of all the forces in the y-direction = 0 AND the sum of all torques about any point = 0

[@thailandfutsal5508]

If sigma F =0 the torque will equal 0 for implicitly wright

[@thailandfutsal5508]

I see ,thank you

[@sanmeetsingh4]

thank you very much .