# Isaac Physics - Pyramid of Spheres

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haven't been able to do part A or C. my working for A leads me to amg/3costheta please help thanks

haven't been able to do part A or C. my working for A leads me to amg/3costheta please help thanks

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(Original post by

Try to find costheta in terms of a and k

**user342**)Try to find costheta in terms of a and k

cos theta = (cos30)a / (1+k) a

what next though?

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#4

(Original post by

i think i have done this -

cos theta = (cos30)a / (1+k) a

what next though?

**jnock04**)i think i have done this -

cos theta = (cos30)a / (1+k) a

what next though?

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(Original post by

Hmm I'm a bit confused by your method. Maybe send a picture of your working?

**user342**)Hmm I'm a bit confused by your method. Maybe send a picture of your working?

ill get a picture of my working in a minute

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#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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#8

(Original post by

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.

**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.

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**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.

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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#10

(Original post by

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:

**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:

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(Original post by

Are you sure it's costheta?

**user342**)Are you sure it's costheta?

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#12

(Original post by

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?

**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?

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(Original post by

Do you know any trig laws, or pythagoras that could help you find sintheta?

**user342**)Do you know any trig laws, or pythagoras that could help you find sintheta?

Last edited by jnock04; 1 month ago

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#14

(Original post by

Yes sorry I wasn't thinking 🤦 find sin theta using sin2theta + cos2theta = 1???

**jnock04**)Yes sorry I wasn't thinking 🤦 find sin theta using sin2theta + cos2theta = 1???

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(Original post by

I personally used pythagoras to find an expression for the opposite edge, but I don't see why that wouldn't work too.

**user342**)I personally used pythagoras to find an expression for the opposite edge, but I don't see why that wouldn't work too.

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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#16

(Original post by

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

**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

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(Original post by

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?

**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?

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#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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(Original post by

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).

**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).

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#20

(Original post by

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?

**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?

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