M3:elastic, Springs And Strings

Can anyone help with these two questions and explain what to do, i think i have a hole in my understanding.

1) Two elastic strings both of natural length 1.5m and moduli 8N and 6N respectively are joined together at both ends. One end of the combined string is fixed and a particle of mass 2kg is attached to the other end and hangs in equilibrium with the string vertical. Calculate the tension in each string.

2) A and B are two fixed points on a smooth horizontal table with AB = 1.5m. A particle is in equilibrium, attached to A by means of a light elastic string of natural length 0.5m and modulus of elasticity lamda N and to B by means of a light elastic string of natural length 0.75m and modulus of elasticity 2lamdaN. Given that the tension in AP is 10N, find the value of lamda.

Please help only 6 days to go until M3 and I'm terrified!!!

Reply 1

dvs

10

1.
Newton's second law:
T₁ + T₂ = 2g

Hooke's law:
(8/1.5)x + (6/1.5)x = 2g, since the extension is the same in both.
x = 2.1

So:
T₁ = (8/1.5) * 2.1 = 11.2N
T₂ = (6/1.5) * 2.1 = 8.4N

2.
T_ap = T_bp, since the particle is in equilibrium.
(k/0.5) x_ap = (2k/0.75) x_bp, by Hooke's law.
6 x_ap = 8 x_bp
x_bp = 0.75 x_ap

We also know that:
0.5 + 0.75 + x_ap + x_bp = 1.5
x_ap + x_bp = 0.25
x_ap + 0.75 x_ap = 0.25
x_ap = 1/7

And we know:
T_ap = 10

So, using Hooke's law:
(k/0.5) x_ap = 10
k = 5/x_ap = 35

Reply 2

KingAS

Lucky Penny

Can anyone help with these two questions and explain what to do, i think i have a hole in my understanding.

1) Two elastic strings both of natural length 1.5m and moduli 8N and 6N respectively are joined together at both ends. One end of the combined string is fixed and a particle of mass 2kg is attached to the other end and hangs in equilibrium with the string vertical. Calculate the tension in each string.

Please help only 6 days to go until M3 and I'm terrified!!!

the two stings are parallel to each other (to help you visulize)... and the first two ends are fixed to and the other two ends are atached to the 2kg weight.

therefore, T1 + T2 = 2g .........1

T1 = [lamda (x)] / l = (8*x)/1.5
T2 = (6*x)/1.5

so, 14x = 29.4
x = 2.1

and T1 = 11.2N
T2 = 8.4 N

im working on the second one now

Reply 3

Lucky Penny

OP

thats great, i think i added an extra tension. Do you assume the extensions are the same because their natural lengths are the same?

Reply 4

dvs

10

I think you misread the question (like I did initially): The strings are attached at both ends.

strings.JPG2.2KB

Reply 5

KingAS

no because the two strings are parallel, the weight is spread between both the strings to make sure the weight is balanced.
hard to explaind without a diagram...

btw i m doin m3 in 6 days too... good luck with ur revision!