Another interesting application of energy is that, since F-_ , we can use the en

Question

Another interesting application of energy is that, since F-_ , we can use the energy of a system to determine the location of an object that makes the entire system stable. That is, when dr = 0, the net force on the system must be zero. Hence, by extremizing the energy, we can then determine the location of an equilibrium position, which can be further analyzed to determine if it is stable or unstable. As an example of this, consider the system of pulleys, masses and string shown in Fig. 1. A light string of length b is attached at point A, passes over a pulley at point B located a distance 2d away, and finally attaches to the mass mi. Another pulley with mass m2 attached passes over the string, fixed a distance c below the pulley, pulling it down between A and B. The pulleys are massless. 2d m1 Figure 1: Cartoon of the pulley problem. 1. Calculate the distance z from point B to the mass m1, when system is in equilibrium Chence, extremize the energy with respect to xi and determine the value for xi). 2. Determine whether the equilibrium is stable or unstable.

Explanation / Answer

1)Let the distance of m1 from B be x.so, length of the string from A to B is b-x.

now,

direct distance between A and B is 2d and m2 is at a distance of D.

Assuming the string to be massless.

Let the tension in the string be T.

balancing the forces on m1,

T-weight = 0

or T = m1*g

let the angle subtended by the thread be y.so,

sin(y) = 2d/(b-x)

now balancing the forces on m2,

2*T*cos(y) = m2*g

or cos(y) = m2/(2*m1)

or 1- ((2d)/(b-x))^2 = (m2/(2*m1))^2

or (2d/(b-x)) = (1-(m2/(2*m1))^2)^0.5

or x = b-2*(d/k)

where k = (1-(m2/(2*m1))^2)^0.5

2)the equilibrium is stable equilibrium since on displacing the system, it returns back to its original position

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