Grade 12 Physical Science: Newton's Laws_3 - Forces acting at an angle

I'm this video, we go through forces in two dimensions and how to tackle questions where a force acts at an angle to the horizontal. Enjoy!

[@itsSARABI]

mr nkosi is going to save my NSC results😭❤

[@MlungisiNkosi]

To great results and beyond

[@glazingglamour]

Facts😂❤

[@KholekaPreet-tw7lo]

You are a life saver Mr Nkosi 🥺🙏❤️,I appreciate you😊

[@sihleyawa7626]

Mr nkosi. a question on your free body diagram for block P are we not supposed to have tension??

[@zeetazerf1247]

that's something I noticed too

[@thando_tlng]

❤Sir uyafundisa iyhoo😮 ande you are patient and you explain in detail 😭❤️ FRIDAY NGIYA PASSA NOMA KANJANI🔥😌

[@StuntmanSkasana]

Yhooo😭🔥the best teacher ever☑️😭🔥🔥

[@MlungisiNkosi]

Thank you 😊

[@talitaallam9379]

sir is not speaking the foreign language of physics but he is speaking english ...... that one that disliked go and argue with your ancestors

[@MlungisiNkosi]

Hahaha... Thank you for coming into my defence. However, I would still want us to create a positive non-threatening environment for everyone, regardless of their opinions. Thank you Talita

[@nangosiyephu339]

Thank you. Keep them. Coming, I'm writing next month

[@malivo_sedi]

Did time table out for next month writing

[@amelanindream]

😭😭😭 consider doing maths 😭too please

[@nangosiyephu339]

@@malivo_sedi I downloaded online with my admission letter so I have everything I need

[@seipatipoopedi1364]

I'm writing on the 04th and 11th of June as well. These videos have been helping with the self-study!

[@MlungisiNkosi]

That is great Seipati. All the best for your exams

[@KgaugeloLesufi-tz8pn]

Sir even if we are given the value of acceleration, but they say forces are moving at constant speed is our fnet still equals to 0? If yes then when are we going to use the acceleration value?

[@kgotsomohale9010]

Mr Nkosi can you pls do a question paper where by we are calculating a frictional force but we are not given coefficient.

[@justsiya7178]

I didn't do well with my Term 1 but I know I will do better with my Grade 12

[@creative_plugsa1073]

For the 5Kg, aren’t we supposed to get the friction to the right and tension going to the left? Tension? String?

[@creative_plugsa1073]

Okay the tension was solved. But isn’t the friction supposed to be on the other side Sir?

[@MlungisiNkosi]

@@creative_plugsa1073 awesome

[@creative_plugsa1073]

Okay I understand. Thank you. Friction is not always opposite the applied force but OPPOSES the motion . Understood

[@MlungisiNkosi]

Yes 😊

[@teressamanzini]

Thank you so much mr Nkosi❤❤😭. Are we allowed to draw the second a free body diagram that includes fx and fy instead of the first one?💙

[@MlungisiNkosi]

Yes you are

[@yamkelayamkela1585]

Sir I am confused as to the statement told us that the constant speed is 2m/s-² and on question 2.3 the speed was zero. In my calculations I included the 2m/s-² and got the Force to be 25,40N. Can you please clarify for me on that🙏

[@MlungisiNkosi]

Be very careful… Constant speed not acceleration

[@creative_plugsa1073]

Sir, Is it wrong to have the friction force oppose the Applied force in almost all cases? Instead of drawing the force to the left I draw it opposite the FApplied?

[@boitumelolesaoana9433]

I did the same

[@MlungisiNkosi]

@@boitumelolesaoana9433 it all depends on the context. There are some scenarios where force and friction are in the same direction

[@happybalu7916]

Sir we don't include -T in diagram for 5kg? For Fnet?

[@othusitseonthusitse3536]

Mr Nkosi l would like to inquire..on the past paper Q2 on 2.3...when drawing full body diagram for force F do we leave out the tension?

[@bontlefakude4331]

If you are drawing a free-body diagram using the components (including Fx and Fy) then you should add Force F

[@jerrydichaba]

This is quite helpful, and thank you. At 01:03:51, when drawing the free body diagram, what should be shown, the force applied or just the vertical and horizontal components, or both?

[@mokgadie_]

You either draw the one with Fx and Fy OR the one with the angled Fa.... They all explain the same thing

[@thandekathusi753]

Thank ye for making physics easier

[@MlungisiNkosi]

Oh my, Oh my Thandeka... It is absolutely a pleasure for me. As you can tell... I absolute love the subject. I'm always trying to find ways to make it easier. I'm glad you're enjoying it

[@zamandlovu3150]

Sir can frictional force magnitude be greater than the applied force? And at what instances does that happen?

[@sphelelengcobo2274]

Obvious yes

[@bafananhlapo7933]

So sir even if the acceleration was given...will it always be equal to 0 if its constant speed?

[@MlungisiNkosi]

That would be a contradiction.

[@siyandadhlamini8826]

@@MlungisiNkosi that's what I'm trying to understand..how come are we using ZERO .. what's the use of 2m•s then

[@zeetazerf1247]

@@siyandadhlamini8826 this is what im questioning too brudda

[@thatomasopha4910]

So sir, how would you do it if you didn't have the 2kg block and you weren't given friction force?

[@MlungisiNkosi]

Now you try it and tell me

[@ibenathibooi5242]

Sir by the last paper 2.4.1) doesn't the tension increase because you decreasing the mass therefore you increasing the acceleration which mean the fnet increases because its directly proportional to the accerleration

[@MlungisiNkosi]

Please provide time stamp of the question you are referring to

[@ibenathibooi5242]

@@MlungisiNkosi 1:13:29

[@sikhonaewert3421]

From the first illustration in the video, if you are only given the weight of the body at rest and coefficient of friction and Force is applied at an unknown angle. What is the the value of F applied in order for the body to move?

[@MlungisiNkosi]

You’d have to calculate the Fx value first and the use Fx = Fcos@ to get the angle

[@khonaniblktv]

Sir at 1:11:30 why is our t value 9.8m besides it being given i wouldve assumed t was unknown

[@moneuamber5948]

Thanks you . Can I ask , for the 5kg block why didn't you count tension when you were writing the equation

[@MlungisiNkosi]

Please provide time stamp

[@moneuamber5948]

Ohh yea sorry about that

[@asandandwandwe6618]

Sir on 5kg block you didn't include tension ??

[@asandandwandwe6618]

Oh sir i got it

[@MlungisiNkosi]

I'm glad you're sorted

[@weightlossmotivation5103]

@@MlungisiNkosi I don't get it

[@hlengi3675]

Sir mina I still don't get it. Why doesn't it have the tension ? 🙏🏽

[@uncleblack6229]

This is really helpful 🤴🔥

[@MlungisiNkosi]

You’re welcome

[@mbathazanele6476]

I used the 2m.s acceleration .got the same answer as your for F =9.24N . But the issues is now My T=7N not 3 N. Please make me understand Malume. At 51:51

[@nyasha0533]

I'm learning alot 😊

[@nyasha0533]

Thank you 😊

[@MlungisiNkosi]

Nothing makes me happier

[@Barbieboyyy]

May God bless you🕯. I owe you one!

[@MlungisiNkosi]

Some day… 🙂

[@lufunogeoffrey2112]

Please help When is tension of both objects equal? And when is tension not equal?

[@MlungisiNkosi]

Tension is equal for both always

[@lufunogeoffrey2112]

@@MlungisiNkosi thank you sir

[@NyefoloMmita]

Goood day I just wanna ask how do you get 9.24 but I get 8.9 can you pls help me understand maybe I'm using my calculator wrong thank you .

[@chair2945]

Thank you so much for this!

[@MlungisiNkosi]

The pleasure is all mine

[@weightlossmotivation5103]

Why did you calculate the kinetic friction using Fy instead of Fx

[@MlungisiNkosi]

Please watch the video before this one

[@weightlossmotivation5103]

@@MlungisiNkosi I watched it

[@MlungisiNkosi]

@@weightlossmotivation5103 my point is that… friction is a function of the normal force, which is a vertical force. You only add force that are in the same dimension (vertical in this case)

[@vusisibeko5497]

Why the acceleration is =0?

[@MlungisiNkosi]

Please give the time stamp

[@m.iichelle.e]

@@MlungisiNkosi 47:54 why is our acceleration 0?

[@NokubongaTsotetsi-f7d]

We are told that the boxes move at a constant speed of 2 m/s why are we say2 (a) is 0

[@silindilevilakazi7142]

Thank you so much sir, it now clear on side 🤗

[@MlungisiNkosi]

You’re welcome ❤️

[@sihleyawa7626]

ohh never mind😁

[@tshegofatsomangaba]

Sir for 2.3 is the acceleration not given as 2

[@tshegofatsomangaba]

Got it 🤦🏾‍♀️the 2 is velocity

[@mpilozuke5248]

Syabonga sir

[@musamhlambi1194]

Thank you so much❤

[@MlungisiNkosi]

You’re welcome

[@mzansi-s]

Good day sir ... There is something that I would like to know.For Question 2 2.3 I did as follows For 20 kg Take right as - Fnet = 0 Fax - T - FF =0 35cos40° - T - 5 = 0 35cos40° - 5 = T......(1) For m T - Fg = 0 T - m(9.8) =0 T=m(9.8) Sub equ 2 in equ 1 35cos40° - 5 = m(9.8) 21,81÷9.8 = m(9.8) ÷ 9.8 m=2.23 kg But then I realised my mistake...but I still have a question,is it possible that the way I substituted can be corrected/marked well of course 😂beside final answer?

[@kwanelengcobo2584]

Asbonge Mlotshwa 🙏🏾

[@rubynaidoo5840]

Hi sir the tension is 32.03 N. You made a mistake with the 4kg block as it was minus fg|| minus ff

[@MlungisiNkosi]

Thank you

[@PIPKILLERS_]

Why did u exclude tension

[@MlungisiNkosi]

It is part of the forces that act on each part

[@weightlossmotivation5103]

@@MlungisiNkosi Can you explain further

[@ChefHot]

👏👏👏👏👏👏

[@MlungisiNkosi]

*take-a-bow*

[@roshaanghani1709]

sir please can you do vectors and working out vectors

[@MlungisiNkosi]

Noted... I will cover that with time. Thank you

[@barhleee1749]

Sir How do we calculate the magnitude of a net force acting on a object

[@MlungisiNkosi]

It depends on what you are given

[@lethaboemily8606]

Thank U sir

[@johnbell1072]

King Nkosi

[@MlungisiNkosi]

🙏🏾

[@lindiwekholopane]

Oh my GOD. I got lost at 46 mins. I don't understand. 😩😩

[@lindiwekholopane]

The Fcos minus 5. Oh MY GOD.

[@lindiwekholopane]

Oh. I.got it.

[@MlungisiNkosi]

I’m glad you eventually got it 😊