# M3 question

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#1
A and B are two fixed points on a smooth horizontal table with AB = 1.5m
A particle P is in equilibrium, attached to A by means of a light elastic strting 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 2 lamda N. Given that the tension in AP is 10N, find the value of lamda.
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#2 0
15 years ago
#3
AP a=(1/2)m, L=LN, x=xm
BP a=(3/4)m, L=2LN x=(0.25-x)m

If the particle is in equilibrium both of the strings have equal tension

T=(Lx)/(a)=10N

10=2Lx (i)
10=(8/3)L(0.25-x) (ii)

(i)=>5=Lx (iii)
(ii)=>(15/4)=0.25L-Lx=>Lx=0.25L-3.75 (iv)

(iii)/(iv)=>0.25L-3.75=5=>0.25L=8.75=>L=35N

Newton.
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