Question

A 2m long light metal rod AB is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One wire is of brass and has cross-sectional area of 0.2 × 10–4 m2 and the other is of steel with 0.1 × 10–4 m2 cross-sectional area. in order to have equal stresses in the two wires, a weight W is hung from the rod. The position of the weight along the rod from end A should be

Answer 1

The weight should be hung at a distance 0.67 m from point A.

Explanation:

The length of rod AB = L=2 m

Suppose a weight W is hung at C at a distance x from A.

• Ts =Tension in steel wire
• Tb =Tension in brass wire

• Stress in steel wire=Ts×As
• Stress in brass wire=Tb×Ab

We are given:

• As = 0.1×10−4m^2
• Ab = 0.2×10−4 m^2

So, for equal stresses

Ts×As = Tb×Ab

Ts×0.1×10^−4 = Tb×0.2×10^−4

Ts/Tb=2 or Ts=2 Tb

Since the system is in equilibrium, the moments of force Ts and Tb about C will be equal.

Thus,Ts×x=Tb(2−x)

2 Tb×x=Tb(2−x)

2x=2−x3x=2

x=2/3 = 0.67 m

Thus the weight should be hung at a distance 0.67 m from point A.

Also learn more

The resistance of a wire is 0.01 cm radius is 10 ohm. If the resistivity of the material of wire 50*10 to the power -8 ohm meter,find the length of the wire.?

https://brainly.in/question/1369563

Answer 2

Answer:

1.33m

Explanation:

t1/a1=t2/a2

t1/0.2*10^-4=t2/0.1*10^-4

t1=2t2

t1*x=2t2(2-x)

x=2(2-x)

x=4-2x

x+2x=4

3x=4

x=4/3

x=1.33m