Competition - post a photo you've taken of nature >>

Forces and strings

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 :frown:

Reply 1

mqb2766

Original post by golgiapparatus31

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

Reply 2

username4407536

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 :/

Reply 3

mqb2766

Original post by golgiapparatus31

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

Reply 4

username4407536

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!

Reply 5

mqb2766

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?

Reply 6

username4407536

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

Reply 7

username4407536

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

(edited 3 years ago)

Reply 8

mqb2766

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.

Reply 9

mqb2766

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