A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.


Now the rod is held by two strings and weight is suspended it from it.


Now the tension in the strings will be in vertically upward direction and the weight of block is acting in vertically downward direction now since rod is held stationary it must be in equilibrium i.e. all the forces on it must cancel each other and also net resultant torque on it due to all the forces must also be zero


Now tension in the strings is the force acting due to them on rod and same amount of force is also acting on them in opposite direction let the Force on Rod due to tension in string A and B be TA and TBrespectively now let the block be at a distance d from string end tied to String A, let the weight of block of mass m be w


Now we know weight is given by


w = m × g


Where m is mass of block and g is acceleration due to gravity


All the forces on the rod is represented by free body diagram as shown in the figure below



So Balancing all the forces for equilibrium condition i.e. net magnitude of forces in vertically upward direction must be equal to magnitude of force in downward direction so we get


TA + TB = w


Now rod is in rotational equilibrium also since it is not rotating so sum of torque due to all the forces acting on rod must be zero around any point


We know magnitude torque on body due to a force with respect to any point is given by



Where r⊥ is the perpendicular distance of point from line of force


And F is the magnitude of force


Now let us consider o point where mass is attached for rotation equilibrium, here forces are perpendicular to rod so the perpendicular distance from any point to line of force is same as the distance from point o now force due to point


We can see torque due to force TA will be in clockwise direction denoted by and due to TBin anticlockwise direction denoted by and there will be no torque to weight of mass m since it is acting on the same point about we are calculating torques, so r⊥ or the perpendicular distance is zero so torque is also zero since


Now the direction of torques are shown in figure below



Now for force TA


r⊥ = d


(Distance from o to left most end where TA is acting)


F = TA


So torque is



Now for force TB


r⊥ = 1.05 – d


(Distance from o to Right most end where TB is acting since total length of rod is 1.05m)


F = TB


So torque is



Now both torque must cancel each other i.e. should be equal in magnitude


i.e.


or


so we get the relation



(a) Now we know stress in a wire is restoring force per unit area , restoring force is same as the Tension in the wire or the force applied by the wire


We know stress in in a wire is given as


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