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Mathematics: Post your doubts here!

usama321

usama321

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x-gamer-x said:
can some one explain me this

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_41.pdf

Q 3

thank u

(MECHANICS 1)
Click to expand...

1.png
NOTE: The tension in the string is equal to T. I am just naming them T1 and T2 so that you can understand them easily, otherwise they are equal.

Resolving forces vertically, there are two forces acting downwards, one is the weight of the ring 8.5N and the other is the component of T2, T2 cos x

One force acting upward which is T1 cos (90-x) which is equal to T1 sin x

thus 8.5 + T2 cosx = T1 sin x

Resolving forces horizontally, there are two forces acting to the left, that are T2 sin x and T1 sin(90-x) which equals T1 cos x.
The other force acting to the right is 15.5 N
Thus
T2 sin x + T1 cosx = 15.5

Now, just put in T for both strings.

Now solve these simultaneously.

T sin x + T cosx = 15.5
T cosx - T sin x = -8.5

2Tcosx = 7
T cos x = 3.5
Similary,
T sin x = 12

Now, rearrange the first equation
x = 3.5/cos x
Put it in the other equation

T sin * 3.5/cos x = 12
3.5T sin x/cos x = 12
sin/ cos = tan
3.5 T tan x = 12

Find the value of x
 
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sitooon

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tompapaya said:
what i would like to add here is that M2 is too difficult for me , so i just choose the S1 to substitute M2. therefore i said "cool"~~
Click to expand...
Do you still have your M2 notes , or know someone who studies M2 .. help i am freaking out this year because of M2:cry:
 
tompapaya

tompapaya

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x-gamer-x

x-gamer-x

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usama321 said:
View attachment 38080
NOTE: The tension in the string is equal to T. I am just naming them T1 and T2 so that you can understand them easily, otherwise they are equal.

Resolving forces vertically, there are two forces acting downwards, one is the weight of the ring 8.5N and the other is the component of T2, T2 cos x

One force acting upward which is T1 cos (90-x) which is equal to T1 sin x

thus 8.5 + T2 cosx = T1 sin x

Resolving forces horizontally, there are two forces acting to the left, that are T2 sin x and T1 sin(90-x) which equals T1 cos x.
The other force acting to the right is 15.5 N
Thus
T2 sin x + T1 cosx = 15.5

Now, just put in T for both strings.

Now solve these simultaneously.

T sin x + T cosx = 15.5
T cosx - T sin x = -8.5

2Tcosx = 7
T cos x = 3.5
Similary,
T sin x = 12

Now, rearrange the first equation
x = 3.5/cos x
Put it in the other equation

T sin * 3.5/cos x = 12
3.5T sin x/cos x = 12
sin/ cos = tan
3.5 T tan x = 12

Find the value of x
Click to expand...
thanks brother
really helped a lot

image.png
 
midha.ch

midha.ch

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Muskan Achhpilia said:
I have a doubt in the following question can someone please helpView attachment 38122

The answer is also given.

Thanks a tonne!
Click to expand...
when line and curve intersects
equation of curve = equation of line
x^2 - 2x = kx - 4
x^2 +(-k-2)x + 4 = 0
a=1, b= -k-2, c=4
given, line intersects curve at 2 distinct points
therefore, b^2 -4ac >0
after solving the inequality, take the 2 critical values of x
write the possible ranges
find the value of individual term in that range
eg- (k+6) is -ve when k<-6
then multiply the signs for each possible range
eg- (k+6)(k-2) = -ve multiplied by -ve = +ve
our initial inequality was greater than 0
thus the range is k<-6 or k>2
 

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S

sitooon

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midha.ch said:
when line and curve intersects
equation of curve = equation of line
x^2 - 2x = kx - 4
x^2 +(-k-2)x + 4 = 0
a=1, b= -k-2, c=4
given, line intersects curve at 2 distinct points
therefore, b^2 -4ac >0
after solving the inequality, take the 2 critical values of x
write the possible ranges
find the value of individual term in that range
eg- (k+6) is -ve when k<-6
then multiply the signs for each possible range
eg- (k+6)(k-2) = -ve multiplied by -ve = +ve
our initial inequality was greater than 0
thus the range is k<-6 or k>2
Click to expand...
Got it , !
We assumed tan(A) = t
so from our t value . we get theta as tan-1(t) > substitut value of t from itertion
 
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