golgiapparatus31
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#1
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#1
https://imgur.com/a/BovGwNR

I'm having problems with 7(b) and 9.

For 7(a) I managed to show given result.

Now for the part (b):
I did this:

Tcos60 = 3.5g
so T = 7g

now X = Tsin60 = 7sqrt(3)/2 g but answer is 7 sqrt(3) g
I am off by a factor of 2 but I'm not able to find why

For 9.
I wrote mgsin25 = Pcos25 + T
so T = 3.9208...
writing T=lambda x / l
I get x/l = 0.392
but I know l+x = 1.5 so I solve for l, getting l = 1.08m

This is correct but I'm not able to get the other answer, 1.22m. No idea how to proceed.

Tried to start by assuming spring is in compression, so T + mgsin25 = Pcos25, but then T is negative, so T must be in tension, so I'm not finding the other possibility
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mqb2766
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#2
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#2
(Original post by golgiapparatus31)
https://imgur.com/a/BovGwNR

I'm having problems with 7(b) and 9.

For 7(a) I managed to show given result.

Now for the part (b):
I did this:

Tcos60 = 3.5g
so T = 7g

now X = Tsin60 = 7sqrt(3)/2 g but answer is 7 sqrt(3) g
I am off by a factor of 2 but I'm not able to find why

For 9.
I wrote mgsin25 = Pcos25 + T
so T = 3.9208...
writing T=lambda x / l
I get x/l = 0.392
but I know l+x = 1.5 so I solve for l, getting l = 1.08m

This is correct but I'm not able to get the other answer, 1.22m. No idea how to proceed.

Tried to start by assuming spring is in compression, so T + mgsin25 = Pcos25, but then T is negative, so T must be in tension, so I'm not finding the other possibility
If you draw the spring triangle and work tension out via extensions you'd get T=14g.
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golgiapparatus31
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(Original post by mqb2766)
If you draw the spring triangle and work tension out via extensions you'd get T=14g.
Spring triangle you mean triangle of forces?
I tried it but didn't get 14g

horizontal side = X
vertically downwards = weight = 3.5g
then tension goes back to complete the triangle
angle between 3.5g side and tension side is 60 degrees
so cos 60 = 3.5g/T
T= 7g

Tried to use extension
(1.2+x)cos60 = 1.2
so x = 1.2
T = 7g from T = lambda x/l

I don't understand :/
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mqb2766
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(Original post by golgiapparatus31)
Spring triangle you mean triangle of forces?
I tried it but didn't get 14g

horizontal side = X
vertically downwards = weight = 3.5g
then tension goes back to complete the triangle
angle between 3.5g side and tension side is 60 degrees
so cos 60 = 3.5g/T
T= 7g

Tried to use extension
(1.2+x)cos60 = 1.2
so x = 1.2
T = 7g from T = lambda x/l

I don't understand :/
Based on lengths, the vertical length is 1.8 (1.2+0.6).
So the hypotenuse is 3.6 (1.2+2.4).
So the tension is 7g*2.4/1.2
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golgiapparatus31
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(Original post by mqb2766)
Based on lengths, the vertical length is 1.8 (1.2+0.6).
So the hypotenuse is 3.6 (1.2+2.4).
So the tension is 7g*2.4/1.2
Ohh okay!! I got it

I made mistake in the equation (1.2+x)cos60 = 1.2. RHS should be 1.8 and it works out

Thanks!
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mqb2766
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(Original post by golgiapparatus31)
Ohh okay!! I got it

I made mistake in the equation (1.2+x)cos60 = 1.2. RHS should be 1.8 and it works out

Thanks!
Are you ok with the other part?
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golgiapparatus31
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(Original post by mqb2766)
Are you ok with the other part?
No. I don't know how to get the other value of 1.22m

Don't know how to start the calculation either
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golgiapparatus31
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(Original post by mqb2766)
Are you ok with the other part?
Hello, I tried using opposite direction of P. It is still a possible situation (reaction = 16.0 > 0) so still in contact with slope
Got tension = 12.98...
x/l = 1.298....
l = 0.653 m

But this value is not 1.22m. Not sure how to get it.

For that value I try to work back. I calculate tension as 2.295... But I don't see how it is possible to get this value with numbers in the question

edit: wrong calc for tension
Last edited by golgiapparatus31; 3 weeks ago
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mqb2766
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(Original post by golgiapparatus31)
No. I don't know how to get the other value of 1.22m

Don't know how to start the calculation either
I can't get it either. If P was pointing out the slope then youd get a natural length of 0.65.
Can't see any other reason.
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mqb2766
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#10
(Original post by golgiapparatus31)
Hello, I tried using opposite direction of P. It is still a possible situation (reaction = 16.0 > 0) so still in contact with slope
Got tension = 12.98...
x/l = 1.298....
l = 0.653 m

But this value is not 1.22m. Not sure how to get it.

For that value I try to work back. I calculate tension as 43.6 N. But I don't see how it is possible to get this value with numbers in the question
We've been trying the same thing. I'd put it in the dodgy question pot and move on ...
The 43.6 doesn't seem right, but it's irrelevant.
Last edited by mqb2766; 3 weeks ago
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golgiapparatus31
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(Original post by mqb2766)
We've been trying the same thing. I'd put it in the dodgy question pot and move on ...
(Original post by mqb2766)
I can't get it either. If P was pointing out the slope then youd get a natural length of 0.65.
Can't see any other reason.
Thank you soo much! You've been a lifesaver!!!
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