An object is attached to four springs. One end of each spring is attached to the

Question

An object is attached to four springs. One end of each spring is attached to the object, while the other is anchored to a point in the x-y plane. The natural lengths of all springs are zero, i.e. the magnitude of the force generated  by the ith spring is given by:

where li is the length of the ith spring. The anchor points and spring constants of the four springs are:

What are the coordinates (x,y) of the system's equilibrium point? [in meters]  

The system is in equilibrium if the mass is located at (x = m, y = m).

Spring constants and anchoring points Spring # ki [N/m] (xi , yi) [m] 1 0.7 (-9, 8) 2 4.0 (4, 4) 3 1.4 (2, 8) 4 3.3 (-4, 5) An object is attached to four springs. One end of each spring is attached to the object, while the other is anchored to a point in the x-y plane. The natural lengths of all springs are zero, i.e. the magnitude of the force generated by the ith spring is given by: where li is the length of the ith spring. The anchor points and spring constants of the four springs are: What are the coordinates (x, y) of the system's equilibrium point? [in meters] The system is in equilibrium if the mass is located at (x =

Explanation / Answer

Let (x,y) be the position of the body at equilibrium,

Then Total force on the body in the x-direction =

Fx = 0.7*[x-(-9)] + 4*(x-4) +1.4*(x-2) + 3.3*[x-(-4)] (since Fx = k*x)

But at equilibrium, Fx = 0

or 0.7*[x-(-9)] + 4*(x-4) +1.4*(x-2) + 3.3*[x-(-4)] = 0

9.4*x = -0.7, x = - 0.0745 m

Total force on the body in the y-direction =

Fy = 0.7*[y-8] + 4*(y-4) +1.4*(y-8) + 3.3*[y - 5] (since Fy = k*y)

But at equilibrium, Fy = 0

0.7*[y-8] + 4*(y-4) +1.4*(y-8) + 3.3*[y - 5] = 0

9.4*y = 49.3, y = 5.245 m

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