Question

A 1.05 m long rod whose weight is negligible is supported at its end by wire A
and B of equal length. The cross-section area of A is 1.00 mm2; that of B is 4.00
mm2. Young’s modulus of wire A is 2.40 1011 a  P ; that for B is1.60 1011 a  P . At
what point along the rod should a weight W be suspended to produce (a) equal
stress in A and B? (c) Equal strain in A and B?
8)

Answer 1

Answer on Question #49488-Physics-Mechanics-Kinematics-Dynamics

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 $1 \, \text{mm}^2$ and that of B is $4 \, \text{mm}^2$ . Young's Modulus for wire A is $2.4 \cdot 10^{11} \, \text{Pa}$ ; that for B is $1.6 \cdot 10^{11} \, \text{Pa}$ . At what point along the rod should weight w be suspended to produce

a) equal stresses in A and B?

b) equal strains in A and B?

Solution

First find tensions in each wire:

$$\sum f = 0: T _ {A} + T _ {B} - w = 0 \rightarrow T _ {A} = w - T _ {B}.$$ $$\sum \tau = 0: (u s e p o s i t i o n o f w a s p i v o t)$$ $$- T _ {A} x + T _ {B} (l - x) = 0$$ $$T _ {B} (l - x) = (w - T _ {B}) x$$ $$l T _ {B} = w x \rightarrow T _ {B} = \frac {w x}{l}; T _ {A} = w - \frac {w x}{l}.$$

a) equal stresses in A and B:

$$\left(\frac {F}{A}\right) _ {A} = \left(\frac {F}{A}\right) _ {B}.$$ $$\frac {w - \frac {w x}{l}}{1 m m ^ {2}} = \frac {\frac {w x}{l}}{4 m m ^ {2}} \rightarrow 1 - \frac {x}{l} = \frac {1}{4} \frac {x}{l} \rightarrow x = \frac {4}{5} l = 1. 0 5 \cdot 0. 8 = 0. 8 4 \mathrm {m f r o m A}.$$

b) equal strains in A and B:

$$\frac {\Delta L}{L _ {0}} = \frac {\left(\frac {F}{A}\right)}{Y}.$$ $$\frac {\left(\frac {F}{A}\right) _ {A}}{Y _ {A}} = \frac {\left(\frac {F}{A}\right) _ {B}}{Y _ {B}}.$$ $$\frac {w - \frac {w x}{l}}{1 m m ^ {2} \cdot 2 . 4 \cdot 1 0 ^ {1 1} P a} = \frac {\frac {w x}{l}}{4 m m ^ {2} \cdot 1 . 6 \cdot 1 0 ^ {1 1} P a} 1 - \frac {x}{l} = \frac {2 . 4}{1 . 6 \cdot 4} \frac {x}{l} \rightarrow x = \frac {3}{8} l = 1. 0 5 \cdot \frac {3}{8} = 0. 3 9 \mathrm {m f r o m A}.$$

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