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# The Official Edexcel M2 Revision Thread! (24/06/05 AM) watch

1. i didnt realise we needed c3 understanding to do m2, i thought it was upto c2, and im fine, but ive only done little bits of p2 and p2 n my spare time. anyone know what trig identities are given to us? i did a double angle question earlier. do you think they will get us to intergrate trig functions?
2. Integrate trig? Probably not...I expect some trigonometry to appear in a projectile question, and then it is just the simplier trig identities..
3. anyone care to help me with a previous post?
http://thestudentroom.co.uk/showpost...6&postcount=18

thanks
4. (Original post by Zuber)
anyone care to help me with a previous post?
http://thestudentroom.co.uk/showpost...6&postcount=18

thanks
well u can still work in the order of the question, by using 25° in the first bit to show that x < 2.5 as in attached diagram
so the line of action of force is still through the base n lamina will b in equilibrium
Attached Images

5. (Original post by hajira)
well u can still work in the order of the question, by using 25° in the first bit to show that x < 5 as in attached diagram
so the line of action of force is still through the base n lamina will b in equilibrium
i thought about that, but thats really jsut the opposite of finding the maximum angle of inclination...kind of (if you know what i mean). seems pointless to have both parts.
6. (Original post by Zuber)
i thought about that, but thats really jsut the opposite of finding the maximum angle of inclination...kind of (if you know what i mean). seems pointless to have both parts.
i think thats wat they r looking for tho
i dont see any other way at least!
anyway..good luck
im off for my S2 exam!
7. plzzzzzzzzzzzz any one had M2 jan03 and june 04
mark scheme
8. http://math.mdsalih.com/data.php?d=E...%20Mathematics

i got jan 03 from there
9. can somebodyu help me?!?! i dont understand centre of mass!!!!!! my teacher taught me to take the masses and the centre mass of each thingie (eg triangle) then do the x(dash) = x1m1 + x2m2.../masses

is that right?!
and i dont understand how i can use moments in centre of masses?!?!!? plzzzzzzzzzzzz help!!!!!!!

thanks guys
10. (Original post by ABUSHOMAL///M3)
plzzzzzzzzzzzz any one had M2 jan03 and june 04
mark scheme
i'v got em both, pm me ur email address n i'll send them
11. thx i`ve brought them
12. (Original post by ossoss87)
can somebodyu help me?!?! i dont understand centre of mass!!!!!! my teacher taught me to take the masses and the centre mass of each thingie (eg triangle) then do the x(dash) = x1m1 + x2m2.../masses
Yeah, from what i understand about centre of mass is that sometimes you have to find the centre of a mass of lamina and so you would split it above as you said there

Sometimes a shape is taken out of another... e.g. a lamina created by a triangle being taken out from a square

Sometimes you have to find the centre of mass assuming the shape is made out of many rods and to do this you use the centre of masses of all the rods

Sometimes there is particles thrown in these centre of mass questions

is that right?!
and i dont understand how i can use moments in centre of masses?!?!!? plzzzzzzzzzzzz help!!!!!!!

thanks guys
ok... you assume that the entire weight of the lamina acts from the point of the centre of mass.

this means that if the lamina is hung from it's corner, for example, there will be a turning force (moment) acting upon it

if you have this lamina in equilibrium you can use the formula

total moments clockwise = total moments anticlockwise

but remember to take the moments from the point that the lamina is hung from

Hope that helps

(and fingers crossed that the june 2005 paper will have no 'find the range of value of e' question even though i'm sure they will)
13. thanks alot revenged!!!!!!!!!!!!!

that really helped...

about the range of values of e.. it will come! definitly... u know with practice u can do them so quick!! if u want help lemme know...

btw guys.. what do u think the moments question will be ?! a ladder?!?!?! (jan 05 tension)
14. (Original post by ossoss87)
thanks alot revenged!!!!!!!!!!!!!

that really helped...

about the range of values of e.. it will come! definitly... u know with practice u can do them so quick!! if u want help lemme know...

btw guys.. what do u think the moments question will be ?! a ladder?!?!?! (jan 05 tension)
i hope its a ladder, i love those questions
15. i hope its a ladder, i love those questions
16. Hey guys this question is confusing the hell out of me. Any solutions or help would be great.

ABCD is a uniform square lamina of side 4m and a mass of 8 kg. It is hinged at A so that it is free to move in a vertical plane. It is maintained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons, Find:
(A) the value of F
(B) the magnitude and direction of the force exerted by the hinge on the lamina.
I'm struggling with A, could someone go through it please.
17. (Original post by k10k)
Hey guys this question is confusing the hell out of me. Any solutions or help would be great.

ABCD is a uniform square lamina of side 4m and a mass of 8 kg. It is hinged at A so that it is free to move in a vertical plane. It is maintained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons, Find:
(A) the value of F
(B) the magnitude and direction of the force exerted by the hinge on the lamina.
I'm struggling with A, could someone go through it please.
Well if I've got the drawing below correct then:

a) you just need to take moments about A. You should get 8gcos45 * √32/2 (from the CofM) = 4F (from D) + √32F cos45 (from C). Just sort out the numbers and I got F = 19.6N

b) The force at A has to act towards the top left, away from the C of M. Resolve the forces in that direction and you get Fa + Fsin45 (from D) = 8gsin45 (from CofM) + Fsin45 (from C) C and D cancel out and Fa = 8gsin45 or 55.44N

Hopefully that's correct and makes sense
Attached Images

18. Nevermind me!!! gonna try it again...

Got same as Zuber....for some reason I drew the horizontal force coming out of B, O__O
19. (Original post by k10k)
Hey guys this question is confusing the hell out of me. Any solutions or help would be great.

ABCD is a uniform square lamina of side 4m and a mass of 8 kg. It is hinged at A so that it is free to move in a vertical plane. It is maintained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons, Find:
(A) the value of F
(B) the magnitude and direction of the force exerted by the hinge on the lamina.
I'm struggling with A, could someone go through it please.
this is so much easier than everyone thinks!! you dont even need to worry about any angles

taking moments around A:

4xF + 4xF = 2x8g
8F = 16g
F=2g= 19.6N

THEN, resolving horizontally (from left to right):
Ag-2g=0
Ag = 2g

THEN vertically (from top to bottom):
Ag+2g-8g=0
Ag = 6g

so the forces at A are 2g (to the left) and 6g (upwards)
overall:
√[(2g)²+(6g)²]=62.0
tan x = 6g/2g
tan x = 3
x = 71.6 degrees
20. Why 4F +4F=2*8g

I understand why there would be one but not two.

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