A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-sectional area 0.1 cm2 and the other of brass ofcross-sectional area 0.2 cm2. Along the rod the distance at which weight be hung to produce equal strains in both the wires is n m, where n is. Ybrass=10 x 1011 dyne/cm2 Ysteel=20 x 1011 dyne/cm2.
Question
A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-sectional area 0.1 cm2 and the other of brass ofcross-sectional area 0.2 cm2. Along the rod the distance at which weight be hung to produce equal strains in both the wires is n m, where n is. Ybrass=10 x 1011 dyne/cm2 Ysteel=20 x 1011 dyne/cm2.
Solution
The problem is asking for the distance at which a weight should be hung on the rod to produce equal strains in both the steel and brass wires.
Strain is defined as the change in length per unit length and is given by the formula:
Strain = Stress/Young's Modulus
Where stress is the force per unit area. In this case, the stress in each wire is the weight of the rod and the hanging weight divided by the cross-sectional area of the wire.
Since we want the strains to be equal, we can set the strains in the two wires equal to each other and solve for the distance at which the weight should be hung.
Let's denote the weight of the rod and the hanging weight as W, the distance from the steel wire at which the weight is hung as x, and the length of the rod as L.
The stress in the steel wire is then Wx/L divided by the cross-sectional area of the steel wire, and the stress in the brass wire is W(L-x)/L divided by the cross-sectional area of the brass wire.
Setting these equal to each other gives:
(Wx/L)/Asteel = (W(L-x)/L)/Abrass
Solving this equation for x gives the distance at which the weight should be hung to produce equal strains in the two wires.
Given that the Young's modulus of brass is half that of steel, and the cross-sectional area of the brass wire is twice that of the steel wire, the weight should be hung at 2/3 the length of the rod from the steel wire, or approximately 133.33 cm.
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