Q. 184.3( 12 Votes )

# 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 mm^{2} and 2.0 mm^{2}, 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.

Answer :

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 T_{A} and T** _{B}**respectively 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

T_{A} + T** _{B} =** 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 T_{A} will be in clockwise direction denoted by and due to T** _{B}**in 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 T_{A}

r_{⊥} = d

(Distance from o to left most end where T_{A} is acting)

F = T_{A}

So torque is

Now for force T_{B}

r_{⊥} = 1.05 – d

(Distance from o to Right most end where T_{B} is acting since total length of rod is 1.05m)

F = T_{B}

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

𝝈 = F/A

Where 𝝈 is the stress in wire, F is the restoring force applied by wire same as applied on it, and A is the area of cross section of the wire

Now here we are given area of cross section of wire A as

A_{A} = 1.0 mm^{2}

Force on wire A is T_{A}

So stress on wire A is given by

𝝈_{A} = T_{A}/A_{A}

And we are given area of cross section of wire B as

A_{B} = 2.0 mm^{2}

Force on wire B is T_{B}

So stress on wire B is given by

𝝈_{B} = T_{B}/A_{B}

Now stress in both the wires is equal

i.e. 𝝈_{A} = 𝝈_{B}

or T_{A}/A_{A} = T_{B}/A_{B}

re arranging we get

T_{A}/T_{B} = A_{A}/A_{B}

i.e. Putting the value of A_{A} and A_{B} we get

or

__Note :____We have not converted area of cross section of wires to m ^{2} from mm^{2} because we were taking their ratio or were dividing them so units would ultimately cancel out , even if we would have converted result would have been same .__

we already know

Where d is the distance of leftmost point to point o where the mass m is attached

Equating both the equation we get

Solving we get

d = 2.1m - 2d

Or 3d = 2.1m

So we have

that means wire should be tied at a distance of 0.7m from the left end of the rod

(b) Now we are given that strain in both the wires A and B must be equal we know strain is ratio of change in length to original length of wire , S = Δl/l where l is original length of wire and Δl is change in length of wire due to application of strain , to find strain in wire we use young modulus which is the ratio of stress to strain and is different for different materials of wire and is given by

Y = Stress/Strain

Or we can write

Strain = Stress/Y

Or we can say

S = 𝝈/Y

Where S is the strain in wire, 𝝈 is the stress and Y is the young’s modulus of wire

Now we know stress in in a wire is given as

𝝈 = F/A

Where 𝝈 is the stress in wire,

F is the restoring force applied by wire same as applied on it, and

A is the area of cross section of the wire

Putting value in S = 𝝈/Y

We get stress , as S = F/AY

Now for wire A

Its made up of steel so young’s modulus of wire is

⇒ Y_{A} = 2 × 10^{11} Nm^{-2}

And stress is ⇒ S_{A} = T_{A}/A_{A}Y_{A}

Where S_{A} is the stress in wire A, A_{A} is area of cross section of wire , T_{A} is the force acting on it and A_{A} is the area of cross section of wire

Now for wire B

Its made up of Aluminium so young’s modulus of wire is

Y_{B} = 7 × 10^{10} Nm^{-2}

And stress is

S_{B} = F_{B}/A_{B}Y_{B}

Where S_{B} is the stress in wire B, A_{B} is area of cross section of wire , T_{B} is the force acting on it and A_{B} is the area of cross section of wire

Since strain in both the wires must be equal

So we have

S_{A} = S_{B}

T_{A}/A_{A}Y_{A} = T_{B}/A_{B}Y_{B}

On rearranging equation we get ratio of forces in two wires as

T_{A}/T_{B} = A_{A}Y_{A}/A_{B}Y_{B}

Putting values of A_{A}, A_{B}, Y_{A}, Y_{B} We get

__Note :____We have not converted area of cross section of wires to m ^{2} from mm^{2} because we were taking their ratio or were dividing them so units would ultimately cancel out , even if we would have converted result would have been same .__

we already know

Where d is the distance of leftmost point to point o where the mass m is attached

Equating both the equation we get

Solving we get

10d = 7.35m - 7d

Or, 17d = 7.35m

So we get

so the mass m must be tied at a distance of 0.43m from the leftmost portion of rod in order to have an equal strain in both the wires

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