91影视

The areas of faces of the cylinders are related by ${\mathrm{A}}_{\mathrm{A}}=2{\mathrm{A}}_{\mathrm{B}}$ and Young moduli of the cylinders by ${\mathrm{E}}_{\mathrm{A}}=2{\mathrm{E}}_{\mathrm{B}}$

Using formula for Young鈥檚 modulus and torque, we can find the magnitude of the forces on the log from wire Aandwire Band theratio ${\mathbf{d}}_{\mathbf{A}}\mathbf{/}{\mathbf{d}}_{\mathbf{B}}$respectively.

Formula:

$\mathrm{E}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{I}/\mathrm{L}}$ 鈥�.. (i)

Here, E is Young鈥檚 modulus, F is force, A is area, I is change in length, and L is original length.

Consider the formula for the torque:

data-custom-editor="chemistry" $\mathrm{味}=\mathrm{Fd}$ 鈥�.. (ii)

Here, data-custom-editor="chemistry" $\mathrm{味}$is torque, F is force, d is perpendicular distance.

From equation (i) for cylinder A and solve as:

${\mathrm{I}}_{\mathrm{A}}=\frac{{\mathrm{F}}_{\mathrm{A}}{\mathrm{L}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{A}}{\mathrm{E}}_{\mathrm{A}}}$ 鈥︹�� (iii)

Similarly, for cylinder B solve as:

data-custom-editor="chemistry" ${\mathrm{I}}_{\mathrm{B}}=\frac{{\mathrm{F}}_{\mathrm{B}}{\mathrm{L}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{B}}{\mathrm{E}}_{\mathrm{B}}}$ 鈥︹�� (iv)

The change in the length of both cylinders is the same. Therefore, equate equations (iii) and (iv) and simplify them further as,

$\begin{array}{rcl}\frac{{\mathrm{F}}_{\mathrm{A}}}{{\mathrm{F}}_{\mathrm{B}}}& =& \frac{{\mathrm{A}}_{\mathrm{A}}{\mathrm{E}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{B}}{\mathrm{E}}_{\mathrm{B}}}\\ & =& \frac{\left(2{\mathrm{A}}_{\mathrm{B}}\right)\left(2{\mathrm{E}}_{\mathrm{B}}\right)}{{\mathrm{A}}_{\mathrm{B}}{\mathrm{E}}_{\mathrm{B}}}\\ & =& 4\end{array}$

Consider the equation as:

data-custom-editor="chemistry" ${\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}=\mathrm{W}$

Solve further as:

$\begin{array}{rcl}\frac{{\mathrm{F}}_{\mathrm{A}}}{\mathrm{W}}& =& \frac{4}{5}\\ & =& 0.80\end{array}$

The fraction of bricks mass supported by cylinder is 0.80

Consider the ratio:

$\begin{array}{rcl}\frac{{\mathrm{F}}_{\mathrm{B}}}{\mathrm{W}}& =& \frac{1}{5}\\ & =& 0.20\end{array}$

The fraction of bricks mass supported by cylinder is 0.20

Applying torque equation about the center of mass is written as follows:

${\mathrm{F}}_{\mathrm{A}}{\mathrm{d}}_{\mathrm{A}}={\mathrm{F}}_{\mathrm{B}}{\mathrm{d}}_{\mathrm{B}}$

Substitute the values and solve as:

$$\begin{gathered} \frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}=\frac{{\mathrm{F}}_{\mathrm{B}}}{{\mathrm{F}}_{\mathrm{A}}} \\ \frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}=\frac{1}{4} \\ \frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}=0.25 \end{gathered}$$N

Therefore, the ratio $\frac{{\mathrm{d}}_{\mathrm{A}}}{{\mathrm{d}}_{\mathrm{B}}}$ is 0.25.