# A-Level Further Maths (Mechanics) question

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#1
A light string has its two ends fastened together to form a loop of natural length 2l and modulus of elasticity mg. The string is placed over two smooth fixed pegs A and B where AB is horizontal and of length 2l. A particle P of mass 2m is attached to the mid-point of the lower part of the loop and hangs in equilibrium vertically below C, the mid-point of AB.

(a) Explain why the extension of the string is equal to twice the length of AP.

(b) Show that the distance CP is l.

I've been staring at this question for a really long time, but I don't get why (a) is true. Please help!
0
17 years ago
#2
Draw a diagram.

a) Let AP = x - then BP = x also, but as AB = 2l, the extension is 2x = 2AP.

b) R(up) [email protected] = 2mg

(HL) => (mgx/l)(CP/x) = mg
=> CP = l
0
17 years ago
#3
(Original post by It'sPhil...)
Draw a diagram.

a) Let AP = x - then BP = x also, but as AB = 2l, the extension is 2x = 2AP.

b) R(up) [email protected] = 2mg

(HL) => (mgx/l)(CP/x) = mg
=> CP = l
0
#4
(Original post by It'sPhil...)
Draw a diagram.

a) Let AP = x - then BP = x also, but as AB = 2l, the extension is 2x = 2AP.

b) R(up) [email protected] = 2mg

(HL) => (mgx/l)(CP/x) = mg
=> CP = l

very much appreciated! thanks!
0
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