A 1001b weight is suspended by three cables as shown in the figure. The ratios o

Question

A 1001b weight is suspended by three cables as shown in the figure. The ratios of the stiffness of the cables are: AP:BP:CP = 2k:k:3k, i.e. cable AP and CP are respectively 2x and 3x the stiffness of cable BP. Before the load is applied, all three cables are unloaded and the geometry is as shown. Assuming the cables are very stiff and the displacements are therefore small, determine the forces in the three cables in terms of stiffness value k and the weight W(= 100). Assuming the springs are not stiff, and therefore the displacements are not necessarily small, develop a set of equations for the displacement of the mass and the forces in the cables.

Explanation / Answer

First part:

As the total weight is supported by three cables, hence the weight must be equal to the sum of the forces on the cables, i.e.

100g = F1 sin 45 + F2 sin 60+ F3sin45

F1 is on AP and lets say displacement in the cable is dx and same is in all the cables because if cables being too stiff. Its force constant is 2k. Hence due to Hooke's law,

F1 = -2k * dx

And its vertical conponent is F1sin45 i.e. -2k *dx* 0.707

Assuming same dispacement in all cables, the force on Bp is given by

F2 = -k * dx

And its vetical component is

F2sin60= -k * dx * 0.5

Similarly,

F3 = -3k * dx

Vertical component is

F3 sin45 = -3k *dx * 0.707

Hence we can find small displacement as

100 g = (-1.414k - 0.5 k - 2.121k)* dx

dx =- 100g/(4.035k)

Hence we can use this value of dx to find F1, F2 and F3

F1 = 2 k * 100g/4.035k

F2= k * 100g/4.035k

F3= 3k* 100g/4.035k

Second part.

The set of equations is given by the equations of force balance in vertical and horizontal directions and by Hooke's law

These are given by:

100g = -2k * dx *sin45 - k*dx*sin60 - 3k*dx*sin 45

0 = -2k*dx*cos45 - k*dx*cos60 + 3k*dx*cos45

And Hooke's law equations as given above.

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