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  1. S

    Mathematics: Post your doubts here!

    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
  2. S

    Mathematics: Post your doubts here!

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

    Mathematics: Post your doubts here!

    Could someone explain why the vertical component of the tention in A pushes upward when the ring is on the point of sliding up in part II
  4. S

    Mathematics: Post your doubts here!

    If you take one angle as a the other angle as 90-a, you get the same answer..
  5. S

    Mathematics: Post your doubts here!

    I was giving an example..you can get a right answer in ALGEBRA through trial and error even though that's not usually the systematic method. N is the contact force/normal reaction. Friction= mule x contact force...they just manipulated the equation
  6. S

    Mathematics: Post your doubts here!

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

    Mathematics: Post your doubts here!

    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
  8. S

    Mathematics: Post your doubts here!

    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?
  9. S

    Mathematics: Post your doubts here!

    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
  10. S

    Mathematics: Post your doubts here!

    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...
  11. S

    Mathematics: Post your doubts here!

    Could someone please explain question 7 to me http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_41.pdf
  12. S

    Mathematics: Post your doubts here!

    If they don't slip, friction is greater than the applied force
  13. S

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    We equate the equations to each other because they're supposed to give the same value of displacement at t=60 right? I just don't understand what the constant represents then? What if they ask us an akin question in the exam, asking us what is denoted by c
  14. S

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    Okay I got it. But in the question. I just posted, in 7(2) theyre assuming that the vertical component of the tention in A is moving up just because the ring is about to move up...but how can that be when the tention in A is pointing toward the ring?
  15. S

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    I don't understand...it will be the constant of the second equation because at time= 60 s, that will be the distance, and the distance after that can be calculated by plugging in the values of x-60 into the integration of the second equation Could you PLEASE do q7
  16. S

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    Why not? I mean it moves that much distance in the first 60 seconds, and then for any time after that, we add the second formula to it by subtracting 60 from it...no? Also could explain q7 from this...
  17. S

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    Can't we just integrate the distance without any limits, putting o as the constant and add the distance moved in the first time interval to the equation formed? Can you please try and see if it works
  18. S

    Mathematics: Post your doubts here!

    Oh wait never mind I get it
  19. S

    Mathematics: Post your doubts here!

    No according to the ms, 'w' is the vertical component of acceleration due to free fall...can that happen...can friction = u x g(cos a)? Why cause g(cos a) isn't the normal reaction :S
  20. S

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    So of we find the distance moved in the first time interval and add it to the equation of the distance moved in the second time interval, would it give us the same resultant equation of the total distance moved?
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