# Isaac Physics - Pyramid of Spheres

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#1
https://isaacphysics.org/questions/s...2020_21_week36

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1 month ago
#2
Try to find costheta in terms of a and k
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#3
(Original post by user342)
Try to find costheta in terms of a and k
i think i have done this -
cos theta = (cos30)a / (1+k) a
what next though?
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1 month ago
#4
(Original post by jnock04)
i think i have done this -
cos theta = (cos30)a / (1+k) a
what next though?
Hmm I'm a bit confused by your method. Maybe send a picture of your working?
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#5
(Original post by user342)
Hmm I'm a bit confused by your method. Maybe send a picture of your working?
wait i did that wrong i think it might be cos theta = a / cos(30) (1+k) a
ill get a picture of my working in a minute
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#6
here
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1 month ago
#7
That's good. You can simplify that expression a bit more, and then extend your method of balancing forces vertically to find R, the reaction force, and balance the components in horizontal direction.
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1 month ago
#8
(Original post by user342)
That's good. You can simplify that expression a bit more, and then extend your method of balancing forces vertically to find R, the reaction force, and balance the components in horizontal direction.
You might need to check when resolving whether it's costheta or sintheta that's needed to find the reaction force.
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#9
(Original post by user342)
That's good. You can simplify that expression a bit more, and then extend your method of balancing forces vertically to find R, the reaction force, and balance the components in horizontal direction.
am i right to say that k^3mg = 3R*cos theta?
if i then substitute my formula for cos theta in i end up with this as the reaction force but i don't think its right:
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1 month ago
#10
(Original post by jnock04)
am i right to say that k^3mg = 3R*cos theta?
if i then substitute my formula for cos theta in i end up with this as the reaction force but i don't think its right:
Are you sure it's costheta?
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#11
(Original post by user342)
Are you sure it's costheta?
Is it sin theta then? If so how would I convert my formula I have for cos theta to sin theta to put it into the tension equation?
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1 month ago
#12
(Original post by jnock04)
Is it sin theta then? If so how would I convert my formula I have for cos theta to sin theta to put it into the tension equation?
Do you know any trig laws, or pythagoras that could help you find sintheta?
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#13
(Original post by user342)
Do you know any trig laws, or pythagoras that could help you find sintheta?
Yes sorry I wasn't thinking 🤦 find sin theta using sin2theta + cos2theta = 1???
Last edited by jnock04; 1 month ago
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1 month ago
#14
(Original post by jnock04)
Yes sorry I wasn't thinking 🤦 find sin theta using sin2theta + cos2theta = 1???
I personally used pythagoras to find an expression for the opposite edge, but I don't see why that wouldn't work too.
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#15
(Original post by user342)
I personally used pythagoras to find an expression for the opposite edge, but I don't see why that wouldn't work too.
tbh that sounds easier i found the opposite edge as 1/2*a + a*sqrt(2k) + ak
not sure if this is right or not because when i divide it by h (a + ak) to find sin theta and then put it in the tension formula from before i still get it wrong. sorry for this bother just really want to figure this out
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1 month ago
#16
(Original post by jnock04)
tbh that sounds easier i found the opposite edge as 1/2*a + a*sqrt(2k) + ak
not sure if this is right or not because when i divide it by h (a + ak) to find sin theta and then put it in the tension formula from before i still get it wrong. sorry for this bother just really want to figure this out
I don't have my working on me atm, but I remember that the opposite edge was something like a*sqrt(k^2+2k-1/3). Do you want to check your working and post it?
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#17
(Original post by user342)
I don't have my working on me atm, but I remember that the opposite edge was something like a*sqrt(k^2+2k-1/3). Do you want to check your working and post it?
this is my working, knowing me i have probably done something stupid with the simple maths
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1 month ago
#18
2 things: check your squaring of a/cos30, and when you're sqrting (op)^2, you've got to sqrt the whole of the rhs. If you sqrt 1 side of an equation, you've got to sqrt the whole of the other side. Eg if I have 5^2=4^2+3^2 (25=16+9), 5 doesn't = 4+3, but it does = sqrt(16+9) <-- sqrt(16+9) doesn't equal sqrt(16)+sqrt(9).
Last edited by user342; 1 month ago
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#19
(Original post by user342)
2 things: check your squaring of a/cos30, and when you're sqrting (op)^2, you've got to sqrt the whole of the rhs. If you sqrt 1 side of an equation, you've got to sqrt the whole of the other side. Eg if I have 5^2=4^2+3^2 (25=16+9), 5 doesn't = 4+3, but it does = sqrt(16+9) <-- sqrt(16+9) doesn't equal sqrt(16)+sqrt(9).
oof so a/cos30 squared should be 4a^2/3 right? then i get the opposite side as sqrt(a(-1/3a +2k + ak^2)). if this is write then it is my working out after that isnt correct now ive fixed this bit?
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1 month ago
#20
(Original post by jnock04)
oof so a/cos30 squared should be 4a^2/3 right? then i get the opposite side as sqrt(a(-1/3a +2k + ak^2)). if this is write then it is my working out after that isnt correct now ive fixed this bit?
Your a/cos(30) looks good to me now. Still not sure on the sqrt. Your method for finding R after seems fine.
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