Equilibrium Problem: Equal Stresses/Strains in A & B

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To achieve equal stresses in wires A and B, the weight should be suspended at the midpoint of the rod, given that wire A has half the cross-sectional area of wire B. For equal strains, the weight must be positioned at a point 1.5 times the length of wire A from the end of the rod, accounting for the different Young's moduli of the wires. The calculations involve using the equations for stress and strain, along with the forces acting on the wires. The problem emphasizes the relationship between tension, cross-sectional area, and Young's modulus in determining the appropriate suspension point. Understanding these principles is crucial for solving equilibrium problems involving different materials.
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Can someone help me with this problem:

A 1.05 m-long rod of negligible weight is supported at its ends by wires A and B of equal length. The cross-sectional area of A is 2.00 mm^2 and that of B is 4.00 mm^2. Young's modulus for wire A is 1.80*10^11 Pa; that for B is 1.20*10^11 Pa. At what point along the rod should a weight w be suspended to produce a) equal stresses in A and B? b) equal strains in A and B?

stress is [pulling force]/[cross-sectional area] and strain is the relative strain of the wires: dL/L. Young's modulus is the ration [stress]/[strain], which is constant.
 
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First find the forces. (Assume the change in geometry due to the strain in the cables is negligible.) For any equations below, assume x = 0 at cable A and x = L at cable B (L = 1.05 m, the length of the rod).

from the sum of forces (in y) = 0:

F_A + F_B = W

where F_A is the tension in cable A, F_B is the tension in cable B, and W is the weight of the object you hang on the rod.

from the sum of moments (about point A in this case, but you can pick another point if you feel like it):

Wx = F_B*L

where x is the position of the hanging weight (distance away from A)

Use those 2 equations to calculate F_A and F_B in terms of W, x, and L.

Now calculate the stress and strain in each cable:

(I assume we're using engineering stress and strain as opposed to true stress and strain. Also, A_A and A_B are the x-sectional areas of cable A and B, respectively.)

σ_A = F_A/A_A
σ_B = F_B/A_B
(and from Hooke's Law)
ε_A = σ_A/E_A
ε_B = σ_B/E_B

(E_A and E_B are the Young's moduli for the two materials, A & B.)

Now you just have to set the quantities the problem asks for equal to each other:

for equal stress:

σ_A = σ_B

for equal strain:

ε_A = ε_B

In each case, substitute in the previous equations so that you can solve for x.


P.S. If anyone reading this knows how to input subscripts and superscripts, please let me know. Thanks.
 


To find the point along the rod where equal stresses are produced in wires A and B, we can use the equation for stress: stress = force/area. Since the length of the wires is the same, we can assume that the force applied on each wire is also equal. This means that the stress in wire A will be equal to the stress in wire B if the cross-sectional area of A is half of the cross-sectional area of B. Therefore, the weight w should be suspended at the midpoint of the rod to produce equal stresses in A and B.

To find the point along the rod where equal strains are produced in wires A and B, we can use the equation for strain: strain = change in length/original length. Since the length of the wires is the same, we can assume that the change in length for each wire is also equal. However, the Young's modulus for wire A is different from that of wire B, which means that the stress in wire A will be different from the stress in wire B if the strains are equal. To achieve equal strains, we need to adjust the weight w so that the stress in wire A is multiplied by the Young's modulus of wire A, which is 1.80*10^11 Pa, and the stress in wire B is multiplied by the Young's modulus of wire B, which is 1.20*10^11 Pa. This means that the weight w should be suspended at a point that is 1.5 times the length of wire A from the end of the rod to produce equal strains in A and B.

In summary, to produce equal stresses in wires A and B, the weight w should be suspended at the midpoint of the rod. To produce equal strains, the weight w should be suspended at a point that is 1.5 times the length of wire A from the end of the rod.
 
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