Solution Found!
Answer: A 1.05-m-long rod of negligible weight is
Chapter 11, Problem 91P(choose chapter or problem)
A \(\text { 1.05-m }\)-long rod of negligible weight is supported at its ends by wires \(A \text { and } B\) of equal length (Fig. P11.91). The cross-sectional area of \(A \text { is } 2.00 \mathrm{~mm}^{2}\) and that of \(B \text { is } 4.00 \mathrm{~mm}^{2}\). Young’s modulus for wire \(A \text { is } 1.80 \times 10^{11} \mathrm{~Pa}\); that for \(B \text { is } 1.20 \times 10^{11} \mathrm{~Pa}\). At what point along the rod should a weight \(w\) be suspended to produce
(a) equal stresses in \(A \text { and } B\) and
(b) equal strains in \(A \text { and } B\)?
Equation Transcription:
Text Transcription:
1.05-m
A and B
A is 2.00 mm^2
B is 4.00 mm^2
A is 1.80 x 10^11 Pa
B is 1.20 x 10^11Pa
w
A and B
A and B
Questions & Answers
QUESTION:
A \(\text { 1.05-m }\)-long rod of negligible weight is supported at its ends by wires \(A \text { and } B\) of equal length (Fig. P11.91). The cross-sectional area of \(A \text { is } 2.00 \mathrm{~mm}^{2}\) and that of \(B \text { is } 4.00 \mathrm{~mm}^{2}\). Young’s modulus for wire \(A \text { is } 1.80 \times 10^{11} \mathrm{~Pa}\); that for \(B \text { is } 1.20 \times 10^{11} \mathrm{~Pa}\). At what point along the rod should a weight \(w\) be suspended to produce
(a) equal stresses in \(A \text { and } B\) and
(b) equal strains in \(A \text { and } B\)?
Equation Transcription:
Text Transcription:
1.05-m
A and B
A is 2.00 mm^2
B is 4.00 mm^2
A is 1.80 x 10^11 Pa
B is 1.20 x 10^11Pa
w
A and B
A and B
ANSWER:
Solution 91P
Step 1 of 8
In the given problem, a rod of negligible weight of length l=1.05 m is suspended using two wires A and B, with cross sectional area and
for wire A and B respectively. Where young’s modulus of wire A is
and for wire B is
. As shown in the figure below,
In part(a) we need to find the distance at which the weight has to be suspended on rod we have equal stresses in A and B. Similarly in part(b) to achieve equal strains in A and B.