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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!!!
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!!!
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.
Tap = Tbp, since the particle is in equilibrium.
(k/0.5) xap = (2k/0.75) xbp, by Hooke's law.
6 xap = 8 xbp
xbp = 0.75 xap
We also know that:
0.5 + 0.75 + xap + xbp = 1.5
xap + xbp = 0.25
xap + 0.75 xap = 0.25
xap = 1/7
And we know:
Tap = 10
So, using Hooke's law:
(k/0.5) xap = 10
k = 5/xap = 35
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.
Tap = Tbp, since the particle is in equilibrium.
(k/0.5) xap = (2k/0.75) xbp, by Hooke's law.
6 xap = 8 xbp
xbp = 0.75 xap
We also know that:
0.5 + 0.75 + xap + xbp = 1.5
xap + xbp = 0.25
xap + 0.75 xap = 0.25
xap = 1/7
And we know:
Tap = 10
So, using Hooke's law:
(k/0.5) xap = 10
k = 5/xap = 35
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!!!
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
thats great, i think i added an extra tension. Do you assume the extensions are the same because their natural lengths are the same?
I think you misread the question (like I did initially): The strings are attached at both ends.
god you're so right i did misread the question... !
Thanks alot to both of you, much appreciated x
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