Question

wo identical elastic cords of negligible relaxed lengths are tied at one of their ends to fixed nails A and B that are equidistant from the origin O. The other ends of the strings are tied to a small ball. To hold the ball in equilibrium at a point P (4 m, 3 m), a force of magnitude F = 1000 N is required. Assuming free space conditions, find force constant of the cords?

Answer 1

let the extended lengths of both be l1 and l2 respectively

forces on string 1 , F1 = k*l1 , and force on string 2 , F2 = is k*l2 , where k is spring constant of these identical strings.

resultant force of F1 and F2 is 100 N

By Parallelogram law of addition of vectors,

100 = sqrt [k^2 ( l1 ^2 + l2 ^2 + 2* l1 * l2 * cosC ] --------------- (1)

where C angle between the strings ( and between F1 and F2)

Let the two pegs be on x-axis at +x and -x , symmetrically either side of origin

CosC = (l1^2+l2^2 - 4x^2 ) / 2*l1*l2 ---- (2) This Cosine Law of Triangles

Substituting value of Cos C from (2) in (1)

100 = k * sqrt ( 2*l1^2 +2*l2^2-4x^2 )

= k sqrt(2) sqrt [ l1^2 +l2^2 - 2x^2] ------- (3)

l1^2 = (4+x)^2 +3^2

l2^2 = ( 4-x)^2 +3^2

Hence

l1^2 +l2^2 - 2x^2 = (4+x)^2 +3^2 + ( 4-x)^2 +3^2 - 2x^2 = 50

Substituting above in (2)

100 = k sqrt ( 2*50)

k = 100/10 = 10 N/m