# Forces and strings

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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

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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#2

(Original post by

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

**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

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(Original post by

If you draw the spring triangle and work tension out via extensions you'd get T=14g.

**mqb2766**)If you draw the spring triangle and work tension out via extensions you'd get T=14g.

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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#4

(Original post by

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 :/

**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 :/

So the hypotenuse is 3.6 (1.2+2.4).

So the tension is 7g*2.4/1.2

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(Original post by

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

**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

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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#6

(Original post by

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!

**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!

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(Original post by

Are you ok with the other part?

**mqb2766**)Are you ok with the other part?

Don't know how to start the calculation either

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(Original post by

Are you ok with the other part?

**mqb2766**)Are you ok with the other part?

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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#9

(Original post by

No. I don't know how to get the other value of 1.22m

Don't know how to start the calculation either

**golgiapparatus31**)No. I don't know how to get the other value of 1.22m

Don't know how to start the calculation either

Can't see any other reason.

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#10

(Original post by

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

**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

The 43.6 doesn't seem right, but it's irrelevant.

Last edited by mqb2766; 3 weeks ago

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(Original post by

We've been trying the same thing. I'd put it in the dodgy question pot and move on ...

**mqb2766**)We've been trying the same thing. I'd put it in the dodgy question pot and move on ...

(Original post by

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.

**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.

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