Question

In the Figure, a lead brick rests horizontally on cylinders A and B. The areas of the top faces of the cylinders are related by A_A=2A_B; the Young’s moduli of the cylinders are related by E_A=2E_B. The cylinders had identical lengths before the brick was placed on them. What fraction of the brick’s mass is supported (a) by cylinder A and (b) by cylinder B? The horizontal distances between the center of mass of the brick and the centerlines of the cylinders are d_A for cylinder A and d_B for cylinder B. (c) What is the ratio d_A/d_B ?

Figure:

Short answer

1. The fraction of bricks mass supported by cylinder is 0.80
2. The fraction of bricks mass supported by cylinder is 0.20
3. The ratio ${\mathrm{d}}_{\mathrm{A}}/{\mathrm{d}}_{\mathrm{B}}$is 0.25

Step-by-step solution

Determine the given quantities

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 ${\mathrm{E}}_{\mathrm{A}}=2{\mathrm{E}}_{\mathrm{B}}$

 Step 2: Determine the concept of torque and Young’s modulus

Using formula for Young’s 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:

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

Here, E is Young’s 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" $\mathrm{\zeta }=\mathrm{Fd}$ ….. (ii)

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

(a) Determine the fraction of brick mass supported by cylinder A

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

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

Similarly, for cylinder B solve as:

data-custom-editor="chemistry" ${\mathrm{I}}_{\mathrm{B}}=\frac{{\mathrm{F}}_{\mathrm{B}}{\mathrm{L}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{B}}{\mathrm{E}}_{\mathrm{B}}}$ …… (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

(b) Determine the fraction of brick mass supported by cylinder B

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

(c) Calculation of horizontal distance between center of mass

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.