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Equilibrium problem

  • Thread starter broegger
  • Start date
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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.
 

jamesrc

Science Advisor
Gold Member
476
1
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.
 

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