Mathematics: Post your doubts here!

[Asad Moosvi]

I

have 3 on 15

Which ones?

 

[Snowysangel]

I did. Got the wrong answer. The question is strange. I mean there is something that is supporting the weight of the ring. And it can't be friction alone, because in the very next question, we use the vertical component of the tension to balance it
Suchal Riaz Saad Mughal Any idea?

But the vertical component of the tension is supposed to point downward according to the head to tail rule

 

[Thought blocker]

Which ones?

Comp paper 1
Eng paper 2
Phy paper 2
:/

 

[Snowysangel]

on the point of sliding upwards means that the friction should act downwards and for the fricton to act downward there must be a force acting upwards which in this case is the vertical component of the tension

But that's not possible! It is implied that the horizontal points to the left...so that the normal reaction can counteract it, but then that would mean that Ta is acting in the same direction as Tc when in effect that's not true

 

[Browny]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w04_qp_4.pdf

Can anybody explain question 6 please in detail? I've been stuck in this question for quite a long while and I would be very thankful to anyone who can give me a detailed explanation if possible?

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_4.pdf

Can anybody explain question 6 please in detail? I've been stuck in this question for quite a long while and I would be very thankful to anyone who can give me a detailed explanation if possible?

Part i )


Part ii) IDK

 

[DeathGamblR]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Hope this helps!

wont it be (8x^((5/2)+1))/(5/2)+1)

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_4.pdf

Can anybody explain question 6 please in detail? I've been stuck in this question for quite a long while and I would be very thankful to anyone who can give me a detailed explanation if possible?

A similar ques was solved just a while ago on the thread... Do search t out. Its well explained

 

[princeali97]

Sorry for the late reply.But here is the solution.
I-
First u can donate the tensions in AB and BC as T1 and T2 respectively.
Second,u can easily calculate the angle BAC and ACB.
BAC=53.13 ACB=36.86.
Now the resolving part.
first resolve vertically you'll get an equation like this:
T1sin53.13 + T2sin36.86=8......(i)
second resolve vertically:
T1cos53.13=T2cos36.86
0.6T1=0.8T2
T1=4/3 T2....(ii)
Equate the value of T1 in equation (i) you will get the value of T2 as 4.8N and then equate the value of T2 in equation (ii) you'll get T1 as 6.4N.
II-
Since equilibrium is limiting and ring is about to slip up the rod,Fmax is in the direction of 2N force (i.e weight of ring)
So, Fmax+2=T1cos53.13
Fmax=6.4cos53.13-2
Fmax=1.84N
uR=1.84 (Fmax=uR) and R=T1sin53.13=5.12N
u(5.12)=1.84
u=0.359.......................
I just want to give you guys an advice.While solving these type of questions u must draw a proper diagram with correct directions of each force.

 

[Ayeshak93]

I have been stuck on this question PLEASE do it someone!!!

And please I need all details, pretend you're teaching a toddler

 

[syed1995]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_43.pdf

Q2 Please.

 

[daredevil]

Sorry for the late reply.But here is the solution.
I-
First u can donate the tensions in AB and BC as T1 and T2 respectively.
Second,u can easily calculate the angle BAC and ACB.
BAC=53.13 ACB=36.86.
Now the resolving part.
first resolve vertically you'll get an equation like this:
T1sin53.13 + T2sin36.86=8......(i)
second resolve vertically:
T1cos53.13=T2cos36.86
0.6T1=0.8T2
T1=4/3 T2....(ii)
Equate the value of T1 in equation (i) you will get the value of T2 as 4.8N and then equate the value of T2 in equation (ii) you'll get T1 as 6.4N.
II-
Since equilibrium is limiting and ring is about to slip up the rod,Fmax is in the direction of 2N force (i.e weight of ring)
So, Fmax+2=T1cos53.13
Fmax=6.4cos53.13-2
Fmax=1.84N
uR=1.84 (Fmax=uR) and R=T1sin53.13=5.12N
u(5.12)=1.84
u=0.359.......................
I just want to give you guys an advice.While solving these type of questions u must draw a proper diagram with correct directions of each force.

Why didnt u take T2 cos into account un the second part?'

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_43.pdf

Q2 Please.

h=0.6 v=4?

 

[Thought blocker]

h=0.6 v=4?

yes

 

[Suchal Riaz]

I am very sory for late response. usama321 daredevil
please look at the diagrams and if it needs explaining tell me i will explain in greater detail.

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_43.pdf

Q2 Please.

here goes what i did
you know that B is going down so
0.6g-T-0.6a-------(1)
T-0.2g=0.2a------(2)
simultaneously equate you get a=5
now use 2as=v^2 where s=1.6 a=5 to find v
then now you know the final speed of A=-4
use 2as=v^2-u^2 to find out s
this will be the extra distance
s+1.6+h=3 sub the values and find it out

 

[A star]

I am very sory for late response. usama321 daredevil
please look at the diagrams and if it needs explaining tell me i will explain in greater detail.

thankyou soooo mush <3 get it after 2 months

 

[Suchal Riaz]

I am almost done with me preparation and i am ready to help. tag me if I unintentionally ignored your question

 

[Suchal Riaz]

thankyou soooo mush <3 get it after 2 months

no problem

 

[Ayeshak93]

View attachment 42224 I have been stuck on this question PLEASE do it someone!!!

And please I need all details, pretend you're teaching a toddler

Suchal Riaz