Question

A horizontal weightless rod of length 3l is suspended on two vertical strings. Two loads of mass m₁ and m₂ are in equilibrium at equal distances from each other and from the ends of the strings (Fig. 29).

Determine the tension T of the left string at the instant when the right string snaps.

Answer 1

At the moment of snapping of the right string, the rod is acted upon by the tension T of

the left string and the forces N₁ and N₂ of normal pressure of the loads of mass m₁ and m₂ (Fig. 165). Since the rod is weightless (its mass is zero), the 

equations of its translatory and rotary motions will have the form

The second equation (the condition of equality to zero of the sum of all moments of force about point O) implies that

N₁ = 2N₂........(1)

Combining these conditions, we get (see Fig. 165)

T = N₁ — N₂ = N₂..............(2)

At the moment of snapping of the right string, the accelerations of the loads of mass m₁ and m₂ will be vertical (point O is stationary, and the rod is inextensible) and connected through the relation

a₂ = 2a₁...............(3)

Let us write the equations of motion for the loads at this instant:

where N'₁ and N'₂ are the normal reactions of the rod on the loads of mass m₁ and m₂. Since N'₁ = N₁ and N'₂ = N₂, we have

Hence we can find N₂, and consequently (see Eq. (2)) the tension of the string