A-Level Maths: R1-12 Forces: Finding the Magnitude and Direction of a Resultant Force

www.buymeacoffee.com/TLMaths
Navigate all of my videos at www.tlmaths.com/
Like my Facebook Page: / tlmaths-1943955188961592 to keep updated
Follow me on Instagram here: / tlmaths

[@Arnenth]

May both sides or your pillow be cold and the blanket cozy and warm. This helped me so so much thank youu

[@WilliamCooper2005]

You're a saint! These videos are very helpful.

[@SadiqunhnabiChoudhury]

It would be very easy to navigate through your videos if you make a seperate playlist for each of the units of a level maths, for e.g. mechanics, statistics and core. Then I'll find it easier to find your videos without having to go through the entire 977 video playlist that you have created on the full a level maths specification to just find a video to help me with mechanics!!!

[@TLMaths]

They're all separated out on my website: sites.google.com/site/tlmaths314/home/a-level-maths-2017/full-a-level

[@SadiqunhnabiChoudhury]

Got it! Thx!

[@Homelander6341]

Helpful videos 💯

[@margheritasacchi8651]

A life savior

[@farahinnajwa4197]

thank you so muchhh

[@xander4003]

At 3:10, I'm getting 8.12N even on degree mode. 7.75N on radian mode. Do you have any idea why?

[@TLMaths]

Yes, I know what you've done wrong. You've found cos(25-9) and sin(25-6) when you should be doing cos(25)-9 and sin(25)-6

[@hisham9041]

king@@TLMaths

[@babajani3569]

Hello sir, when you take a direction as being positive (for either forces or suvat or moments) then does that mean that anything that is in the opposite directing will be negative. Can you be sure of that or fo tou still have to common sense check it? Thank you.

[@TLMaths]

I'm not sure what you would check - it should all work out. If you take a direction to be positive then the opposite direction will be negative.

[@susua7336]

why is the bearing not 360-33.74 degrees which would give 326 ? why have you inverted the triangle when trying to calculate the bearing ?

[@TLMaths]

The vector is -1.749i - 2.619j which points bottom left (south west). A bearing of 326 degrees would be pointing north west

[@nadeen-2867]

if they only ask for the magntiude and direction, do i still have to find the bearing?

[@TLMaths]

You will either be asked for the bearing or the direction. If asked for the direction, they would likely tell you how they want it to be measured (e.g. anticlockwise from the positive horizontal axis)

[@nadeen-2867]

@@TLMaths thank you!

[@mahiraarsalan]

Is this video suitable for CIE?

[@latanyajones]

right now i am studying this and a mind said test myself just to see i could not do it

[@StreamerHighlightsHQ834]

at 5 minutes, why did you times the i and j vector by -1?

[@TLMaths]

8*cos(25)-9 = -1.749... 8*sin(25)-6 = -2.619... So the sides of the right-angled triangle I drew are 1.749... and 2.619...

[@tanyavanwyk8319]

@@TLMaths can you explain why you multiplied by -1 please, as I am still unclear as to why you did that

[@TLMaths]

8*cos(25)-9 = -1.749... so the horizontal length is 1.749... this is the same as -(-1.749...) = -(8*cos(25)-9) = 9-8*cos(25)

[@EB_171]

dont get why you square the component forces

[@EB_171]

or use pythagoras

[@TLMaths]

The magnitude of a vector is finding its length. When you draw a 2D vector, it has a horizontal and vertical component, so you can make a right-angled triangle, and hence use Pythag to find the length.

[@chloedingle9450]

Why do you multiply it by -1

[@TLMaths]

Can you give me a timestamp of when you mean?

[@chloedingle9450]

@@TLMaths Hi sir, it’s at 5:12

[@TLMaths]

Ah I see, so 8cos(25)-9 is negative. When labelling my triangle, I want to use the *lengths* of the sides. The *length* of 8cos(25)-9 is 9-8cos(25), which is the same as multiplying it through by -1. It's like saying that you have -8, but I don't want to label the side of a triangle with -8, I'd prefer to label it with the side's length, which is 8.

[@chloedingle9450]

@@TLMaths Ahh I understand thank you! I have paper 1 on Tuesday and I want to die!

[@Masowe.]

@@chloedingle9450 how did you do

[@ebenezerlombiso3816]

U lost me

[@TLMaths]

Anything I can help clarify?

[@ebenezerlombiso3816]

@@TLMaths STION 2 2.0 A bird has a mass of 26 g and perches in the middle of a stretched telephone line. 2.1 Show that the tension in the line can be calculated using the equation T = mg 2 sin θ . 2.2 Determine the tension when: 1.2.1 θ = 5°. 1.2.2 θ = 0.5°. (Assume that each half of the line is straight.) Can u help me solve this please thank you.

[@TLMaths]

I thought you were going to ask a question about this specific video!! Here's my answer: jamboard.google.com/d/1FimdL2BdhgKLyfx4Z624PVdMt_TWdladtdvw_AJ3qZY/edit?usp=sharing

[@ebenezerlombiso3816]

@@TLMaths thank you I really appreciate it