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Mathematics: Post your doubts here!

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The first inequality actually makes a vertical line at x = 1 right? I think in your diagram you considered 2 to be the y coordinate by mistake. Also, in your diagram you started the arg angle from the origin. I thought I should start the angle line from ( -2, 1 ) and focus it towards the right till it intersects my vertical x line? That triangle would represent my inequality right?

I've attached a sort of rough sketch. Its quite terrible, I know. Is that how its suppose to look and I assume that I should join the coordinate from u till it touches my vertical line to make my base aswell so it can be a complete triangle which represents the shaded area?
Yes, i misread it, you are right
:)
 
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hey could u help me by telling me how to find normal contact force(R) in mechanics...like when a ring is being pulled at some angle from a rod
 
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thnx man but still not very clear with the triangle...like what andle does F(N) make with the rod?by the way could u be clear of your location?

That angle is usually mentioned in the question. Is there any specific mechanics question about this rod-ring thing?
 
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That angle is usually mentioned in the question. Is there any specific mechanics question about this rod-ring thing?
yeah like in oct/nov 09 . i.e. 9709 w09 qp 42 question no 5...i did look at the mark scheme and got it...but i need to understand it for other similar questions like in w06 42 question no 2
 
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hey guys could you help me with understanding the sketching the argand diagram like explain what |z-1-i√3|≤ 1 and arg z ≤ pi/3
 
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yeah like in oct/nov 09 . i.e. 9709 w09 qp 42 question no 5...i did look at the mark scheme and got it...but i need to understand it for other similar questions like in w06 42 question no 2

This's how we'll do Q2, P4 Nov. 06.

Part (i):

Friction1.JPG

We'll be needing to resolve the forces vertically in order to find the magnitude of the frictional force. Note that the vertical component of the 5N force and the frictional force act in the same direction (upwards) while as the weight component acts downwards.

Vertical component of the 5N force + Frictional Force = Weight Component
5 cos 30 + F = 6
F = 6 - 5 cos 30
F = 1.67N

Therefore, Frictional Force = 1.67N.

Part (ii):

Friction.JPG

To find the coefficient of friction between the ring and the rod, we'll use the formula F=uR. We've already calculated the 'F' in first part of this question and in order to calculate the 'Reaction Force (R)', we'll resolve the forces horizontally.

R = 5 sin 30
R = 2.5

F=uR
1.67 = 2.5u
0.668 = u

Therefore, the coefficient of friction between the ring and the rod is '0.668'.
 

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This's how we'll do Q2, P4 Nov. 06.

Part (i):

View attachment 6287

We'll be needing to resolve the forces vertically in order to find the magnitude of the frictional force. Note that the vertical component of the 5N force and the frictional force act in the same direction (upwards) while as the weight component acts downwards.

Vertical component of the 5N force + Frictional Force = Weight Component
5 cos 30 + F = 6
F = 6 - 5 cos 30
F = 1.67N

Therefore, Frictional Force = 1.67N.

Part (ii):

View attachment 6286

To find the coefficient of friction between the ring and the rod, we'll use the formula F=uR. We've already calculated the 'F' in first part of this question and in order to calculate the 'Reaction Force (R)', we'll resolve the forces horizontally.

R = 5 sin 30
R = 2.5

F=uR
1.67 = 2.5u
0.668 = u

Therefore, the coefficient of friction between the ring and the rod is '0.668'.
thanks a lot man!!really!!!
 
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Q. For the GP 4,6,9.... find the value of n if the nth term is the first term greater than 100.
Help me plzz
 
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