Mathematics: Post your doubts here!

[kitkat <3 :P]

Could someone please explain question 7 to me http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_41.pdf

lol m stuck on the same question

 

[Snowysangel]

lol m stuck on the same question

After staring at the question for about an hour I finally understood the first part...but in the second part they're assuming that the vertical component of the tention is pulling upward even though the tention is pointing toward the ring and its vertical conponent is hence supposed to point downward :S

 

[usama321]

Asad Moosvi
They have told us that no sliding takes place between the two blocks. Thus ma must not be greater than the frictional force between the blocks. Note: No force is being applied on block A itself, however when b accelerates, it would accelerate too, and there would be a force on it.
2000*.2
Thus, the force must be
200a <= 2000*.2
a < = 2
This means that if the blocks accelerate more than 2m/s^2, block A will slide
Maximum possible value of P would be when block b has acceleration of 2m/s^2

P - 4500*.7 = 450 *2

 

[usama321]

Could someone please explain question 7 to me http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_41.pdf

I am confused about this question too. The thing that confuses me is why we don't take the weight of the ring in our calculations

 

[kitkat <3 :P]

I am confused about this question too. The thing that confuses me is why we don't take the weight of the ring in our calculations

well maybe because the question says the ring in fixed ._.

 

[usama321]

well maybe because the question says the ring in fixed ._.

It says the rod is fixed not the ring

 

[usama321]

After staring at the question for about an hour I finally understood the first part...but in the second part they're assuming that the vertical component of the tention is pulling upward even though the tention is pointing toward the ring and its vertical conponent is hence supposed to point downward :S

Can you please tell me why don't we take the weight of the ring into considerations while resolving forces vertically?

 

[kitkat <3 :P]

It says the rod is fixed not the ring

ohh acha

 

[Snowysangel]

Can you please tell me why don't we take the weight of the ring into considerations while resolving forces vertically?

It doesn't exert a force ON the string. Think about it in a practical situation. The ring and the rod are just support points...their weights don't affect the tention in the string. The forces exerted by he ring, however, do action the rod

 

[daredevil]

Can you please tell me why don't we take the weight of the ring into considerations while resolving forces vertically?

Snowysangel and kitkat <3 :P

yeaahhh me too!! one of my forces is 5.2N and the other is 6.4 N

I considered the weight of the ring too ....
and i dont think the ring is fixed....

even if so.... how come one of my forcess is accurate while the other isn't?! >_< even thought i used one (THE APPARENTLY WRONG ONE) to calculate the other one (WHICH CAME OUT TO BE CORRECT)

cud it be that the ms is wrong?? has anyone seen if the er has anything to say about this?

 

[kitkat <3 :P]

It doesn't exert a force ON the string. Think about it in a practical situation. The ring and the rod are just support points...their weights don't affect the tention in the string. The forces exerted by he ring, however, do action the rod

i'm not even able to do the first part

 

[Snowysangel]

i'm not even able to do the first part

Lol mark two angles within the triangle, alpha and beta. Alpha is between the hypotenuse (2.5 m) and C and beta is between the hypotenuse and A. Solving the vertical and horizontal components, Tc x cos(alpha) + Ta x cos(beta) = 8 and Tc x sin(alpha) = Ta x sin (beta)...do you get it now?

 

[Thought blocker]

Lol mark two angles within the triangle, alpha and beta. Alpha is between the hypotenuse (2.5 m) and C and beta is between the hypotenuse and A. Solving the vertical and horizontal components, Tc x cos(alpha) + Ta x cos(beta) = 8 and Tc x sin(alpha) = Ta x sin (beta)...do you get it now?

No

 

[Snowysangel]

Snowysangel and kitkat <3 :P

yeaahhh me too!! one of my forces is 5.2N and the other is 6.4 N

I considered the weight of the ring too ....
and i dont think the ring is fixed....

even if so.... how come one of my forcess is accurate while the other isn't?! >_< even thought i used one (THE APPARENTLY WRONG ONE) to calculate the other one (WHICH CAME OUT TO BE CORRECT)

cud it be that the ms is wrong?? has anyone seen if the er has anything to say about this?

As long as you apply logic and it make sense, your answer will always be correct...you probably did something wring with the other one. I mean its fit the same reason that trial and error gives you the correct answer

 

[kitkat <3 :P]

Lol mark two angles within the triangle, alpha and beta. Alpha is between the hypotenuse (2.5 m) and C and beta is between the hypotenuse and A. Solving the vertical and horizontal components, Tc x cos(alpha) + Ta x cos(beta) = 8 and Tc x sin(alpha) = Ta x sin (beta)...do you get it now?

cos alpha=2/2.5
cos beta = 1.5/2.5 right?

 

[Snowysangel]

No

There are multiple questions on the past papers related to this but the angles are usually given to make it easier for us

 

[Thought blocker]

There are multiple questions on the past papers related to this but the angles are usually given to make it easier for us

cos alpha=2/2.5
cos beta = 1.5/2.5 right?

Got it. -Phew- M1 is boring :/

 

[daredevil]

i didnt get wat u just said about trial and error... where are we using trial and error here???

and Why in the HELLO r we using u=F/N
what is THAT?! >__< i swear i dont know wat i;m gonna do in the paper tomorrow!! :O

 

[daredevil]

cos alpha=2/2.5
cos beta = 1.5/2.5 right?

okay u got it... now explain it to me....

i used the whole method like finding the angle BAC then subtracting that from 90 to get the acute angle @A ... u know the one on the outer siide..
then the same whole method with the other one..

i'll try and show u wat i did...

 

[kitkat <3 :P]

okay u got it... now explain it to me....

i used the whole method like finding the angle BAC then subtracting that from 90 to get the acute angle @A ... u know the one on the outer siide..
then the same whole method with the other one..

i'll try and show u wat i did...

i didnt find the angles i just found out the ratios like if m considering angle alpha so cos alpha=Base/hypotenuse = 2/2.5 for hypotenuse apply pythagorus