Please try to make a video on scale pans, lorries, elevators etc connected to a point by a light inextensible string. They've got em questions for the new Mechanics 1 of Edexcel A levels
i know this is message is a year ago but here goes nothing ^^ the 3g has a horizontal and vertical component. Vertical is 3gcos20 and is perpendicular to the plane the 3gsin20 is the horizontal component which acts parallel to the plane. It changes because your resolving vectors in the horizontal direction.
It looks like he multiplies the mass of A by g (6 x g = 6g) which is the force of gravity on A. Then he multiplies that by 0.6 which is the frictional force, making "Fr" (force of resistance I'm guessing) 3.6g. I've never seen a rough plane, so this was a first for me.
@@hudy8867 I’m guessing you already know the answer, but for anybody else The way to find friction using the coefficient of friction is to multiply the coefficient of friction by Fn (normal force), which on an incline would be equal to mgcos*theta as this is equal to the normal force on an incline plane.. on a flat surface, the normal force is equal to Mg, so it would be mg x the coefficient of friction, which is usually given in the question details ..
@k.raini no even if the heavier particle is on the table, it will still be pulled off by the lighter particle. Unless something is pulling the heavier particle away from the edge of the table
@@hayatburak9252 A-Levels and their content will vary based on which exam board you use. The most common ones are the AQA, OCR, Edexcel Exam Boards. Less common ones exist like OCR B (MEI), WJEC ...
thank you so much I understand everything in my maths course so far apart from the mechanics section and I feel much more confident now. :)
at 12:07 how do you know whivh direction it accelerates in
AySantana use practical knowledge 4g is greater that 3(a)
Please try to make a video on scale pans, lorries, elevators etc connected to a point by a light inextensible string. They've got em questions for the new Mechanics 1 of Edexcel A levels
At 15:26 why does it change from 3gCos20 to 3gSin20 ?
i know this is message is a year ago but here goes nothing ^^ the 3g has a horizontal and vertical component. Vertical is 3gcos20 and is perpendicular to the plane the 3gsin20 is the horizontal component which acts parallel to the plane. It changes because your resolving vectors in the horizontal direction.
@@raindrop9314 lol
I know Im late but it's 3gsin20 as this is the horizontal component of weight, which we want to use. 3gcos20 would be the vertical component
@@sameers7461 how do you know when to use cos or sin
@@muffincakes3768 I believed it is with the use of sohcahtoa since this is a triangle
At 6:55 how did u calculate "Fr", i dont get it, why multiply 6g * 0.6
It looks like he multiplies the mass of A by g (6 x g = 6g) which is the force of gravity on A. Then he multiplies that by 0.6 which is the frictional force, making "Fr" (force of resistance I'm guessing) 3.6g. I've never seen a rough plane, so this was a first for me.
@@hudy8867 I’m guessing you already know the answer, but for anybody else
The way to find friction using the coefficient of friction is to multiply the coefficient of friction by Fn (normal force), which on an incline would be equal to mgcos*theta as this is equal to the normal force on an incline plane.. on a flat surface, the normal force is equal to Mg, so it would be mg x the coefficient of friction, which is usually given in the question details ..
@@turkeybacon1199 thankyou youre the best this had me stumped for real
does maths genie make wishes come true?
thanks sir for teaching this lesson
sir, on the second question how do you know which way it accelerates
The system will move in the direction of the heavier particle.
@@imichard4857 but 4kg is less than 6kg? I’m confused
@k.raini no even if the heavier particle is on the table, it will still be pulled off by the lighter particle. Unless something is pulling the heavier particle away from the edge of the table
even if you pick the wrong way, you will get a negative acceleration of the same magnitude
when doing 3gsin20 shouldnt it be (3g)/(sin20)?? Kinda confused?
Anyone if you could help :)
No it’s 3gsin20 because your trying to find the vertical component of weight
Do you get 3gsin20, because we find the adjacent length
where does 0.392g come from?
I know right
Because g = 9.8N and weight = mass * gravity so 0.04g = 0.04(9.8) which is equal to 0.392
@@vvoralia5913 thanks queen
what board is this
Arjun Parmar hello n1gga
I think it can be used for all exam boards
what do you mean by ‘board’ ?
@@hayatburak9252 A-Levels and their content will vary based on which exam board you use. The most common ones are the AQA, OCR, Edexcel Exam Boards. Less common ones exist like OCR B (MEI), WJEC ...
@@hayatburak9252 The same goes for GCSEs ...