In Fig. 12-60, a 103 kguniform log hangs by two steel wires, Aand B, both of radiuses 1.20 mm. Initially, wire Awas 2.50 mlong and 2.0 mmshorter than wire B. The log is now horizontal. What are the magnitudes of the forces on it from (a) wire Aand (b) wire B? (c) What is the ratio dA/dB?

1. The thread breaks under a stress of $8.20\times {10}^8\mathrm{N}/{\mathrm{m}}^2$and strain of 2.00
2. Initial length of 2.00 mm and cross-sectional area$8.00\times {10}^{-12}{\mathrm{m}}^2$

Young’s modulus is equal to stress divided by strain. Stress is equal to force per unit area. Strain is equal to a change in length per unit of the original length.

Using the formula for Young’s modulus and torque, and with the help of static equilibrium condition, we can find the magnitude of the forces on the log from wire Aandwire Band theratio dA/dBrespectively.

Consider the formula for the elastic modulus as:

$\mathrm{E}=\frac{\mathrm{F}/\mathrm{A}}{∆\mathrm{L}/\mathrm{L}}$ ….. (i)

Here, E is Elastic Modulus or Young’s modulus, F is force, A is area,$∆\mathrm{L}$ is change in length L original length.

Consider the formula for the torque as:

$\mathrm{\zeta }=\mathrm{d}\times \mathrm{F}$ ……. (ii)

Here, $\mathrm{\zeta }$is torque, F is force, d is the perpendicular distance from point of rotation to the point of application of force.

Since the log is in equilibrium,

${\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}-\mathrm{mg}=0$

Here, F_A and F_B are the forces exerted by the wires on the log, m is the mass of the log.

Resolve the equation as:

$\mathrm{E}=\frac{{\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}}{{\displaystyle \frac{∆\mathrm{L}}{\mathrm{L}}}}$

For wire A, the change in length is obtained as:
$∆{\mathrm{L}}_{\mathrm{A}}=\frac{{\mathrm{F}}_{\mathrm{A}}{\mathrm{L}}_{\mathrm{A}}}{\mathrm{AE}}$ (iii)

Similarly, for wire B change in the length as:

$∆{\mathrm{L}}_{\mathrm{B}}=\frac{{\mathrm{F}}_{\mathrm{B}}{\mathrm{L}}_{\mathrm{B}}}{\mathrm{AE}}$ (iv)

Consider the l is the amount of length by which wire B is longer than wire A. Both wires have same length after the log is attached.

Rewrite the equation as:

$∆{\mathrm{L}}_{\mathrm{A}}=∆{\mathrm{L}}_{\mathrm{B}}+\mathrm{I}$

Substitute the values from equation (iii) and (iv) and solve as:

$$\begin{gathered} \frac{{\mathrm{F}}_{\mathrm{A}}{\mathrm{L}}_{\mathrm{A}}}{\mathrm{AE}}=\frac{{\mathrm{F}}_{\mathrm{B}}{\mathrm{L}}_{\mathrm{B}}}{\mathrm{AE}}+\mathrm{I} \\ {\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}-\mathrm{mg}=0 \\ {\mathrm{F}}_{\mathrm{A}}=\frac{{\mathrm{mgL}}_{\mathrm{A}}+\mathrm{AEI}}{{\mathrm{L}}_{\mathrm{A}}+{\mathrm{L}}_{\mathrm{B}}} \end{gathered}$$

The cross-sectional area of a wire is calculated as:

$\mathrm{A}={\mathrm{\pi r}}^2$

Substitute the values in the equation of the force and solve as:

The magnitude of the forces on the log from wire A is 866 N

We have,

The magnitude of the forces on the log from wire is

Let’s consider origin on the surface of the log at a point directly above the center of mass. Applying torque at this point, therefore,

.

Therefore, the ratio is