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Q11P

Expert-verified
Found in: Page 207

### Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

# An ${\mathbf{800}}{\mathbf{kg}}$load is suspended as shown in Figure ${\mathbf{5}}{\mathbf{.}}{\mathbf{69}}$. (a) Calculate the tension in all three wires (that is, the magnitude of the tension force exerted by each of these wires). (b) These wires are made of a material whose value for Young’s modulus is ${\mathbf{1}}{\mathbf{.}}{\mathbf{3}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{11}}}{\mathbf{N}}{\mathbf{/}}{{\mathbf{m}}}^{{\mathbf{2}}}$. The diameter of the wires is ${\mathbf{1}}{\mathbf{.}}{\mathbf{1}}{\mathbf{m}}$. What is the strain (fractional stretch) in each wire?

(a) The tension in all of the wires is $7840\mathrm{N},6030\mathrm{N}$and $4020\mathrm{N}$respectively.

(b) The strain in each wire is $0.063,0.048$and $0.032$respectively.

See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as:

• The mass of the load is$m=800\mathrm{kg}$.
• The diameter of the wires is$d=1.1mm×\left(\frac{{10}^{-3}m}{1m}\right)=1.1×{10}^{-3}m$.
• The value of the young’s modulus of the material is$\mathrm{y}=1.3×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}$.

## Step 2: Significance of Newton’s second law and the young’s modulus

Newton’s second law states that the force exerted on an object is equal to the product of the mass and the acceleration of that object.

Young’s modulus is described as the ratio of the stress to the strain. The stress is the force exerted by an object per unit area, and the strain is described as the elongation in length.

## Step 3:- (a) Determination of the tension in the wires

The free-body diagram of the acting forces has been drawn below:

The main forces are ${F}_{1},{F}_{2},{F}_{3}$ and the component of the main forces acting in the $x$direction is and ${\mathrm{F}}_{3\mathrm{x}},\mathrm{and}\mathrm{y}$ direction is ${\mathrm{F}}_{2\mathrm{y}}\mathrm{and}{\mathrm{F}}_{3\mathrm{y}}$.

From Newton’s second law, the equation of the tension force applied in the first wire is expressed as:

${F}_{1}=mg$

Here, $m$is the mass of the load and $g$is the acceleration due to gravity that has the value $9.8m/{s}^{2}$.

Substitute $800\mathrm{kg}$ for$m$ and$9.8m/{s}^{2}$ for $g$.

${F}_{1}=\left(800kg\right)×\left(9.8m/{s}^{2}\right)\phantom{\rule{0ex}{0ex}}=7840kg.m/{s}^{2}\phantom{\rule{0ex}{0ex}}=7840kg.m/{s}^{2}×\frac{1N}{1kg.m/{s}^{2}}\phantom{\rule{0ex}{0ex}}=7840N$

There are two components in the tension force of the second and the third wire are ${\mathrm{F}}_{2\mathrm{y}}\mathrm{and}{\mathrm{F}}_{2\mathrm{x}}\mathrm{and}{\mathrm{F}}_{3\mathrm{x}}\mathrm{and}{\mathrm{F}}_{3\mathrm{y}}$. According to the diagram, the force ${\mathrm{F}}_{2}\mathrm{and}{\mathrm{F}}_{3}$ makes the angle${\mathrm{\theta }}_{1}=60°\mathrm{and}{\mathrm{\theta }}_{2}=40°$ with the $x$axis respectively.

The equation of the change in the momentum with time in the $y$-direction is expressed as:

$\frac{d{\stackrel{\to }{p}}_{y}}{dt}=0\phantom{\rule{0ex}{0ex}}={F}_{net.y}$

Here, $\frac{d{\stackrel{\to }{p}}_{y}}{dt}$ is the change in the momentum and ${F}_{net.y}$is the net force acting in the $y$-direction.

Here, from the above figure, the summation of the forces ${F}_{1},{F}_{2y},{F}_{3y}is0$, then this equation can also be written as:

${F}_{2y}+{F}_{3y}={F}_{1}\phantom{\rule{0ex}{0ex}}{F}_{2}\mathrm{cos}\left(90-{\theta }_{1}\right)+{F}_{3}\mathrm{cos}\left(90-{\theta }_{2}\right)={F}_{1}$ …(i)

Substituting $60°\mathrm{for}{\mathrm{\theta }}_{1}\mathrm{and}40°\mathrm{for}{\mathrm{\theta }}_{2}$ in the above equation.

$0.86{F}_{2}+0.64{F}_{3}=7840$

The equation of the change in the momentum with time in the x-direction is expressed as:

$\frac{d{\stackrel{\to }{p}}_{x}}{dt}=0\phantom{\rule{0ex}{0ex}}={F}_{net.x}$

Here, $\frac{d{\stackrel{\to }{p}}_{y}}{dt}$is the change in the momentum and ${F}_{net.x}$is the net force acting in the $x$-direction

As these two forces ${F}_{3x}$are in opposite directions, then the summation of these forces is equal to $0$.

${F}_{2x}-{F}_{3x}=0\phantom{\rule{0ex}{0ex}}{F}_{2}\mathrm{cos}{\theta }_{1}-{F}_{3}\mathrm{cos}{\theta }_{2}=0\phantom{\rule{0ex}{0ex}}{F}_{2}={F}_{3}\frac{\mathrm{cos}{\theta }_{2}}{\mathrm{cos}{\theta }_{1}}$

Substituting $60°\mathrm{for}{\mathrm{\theta }}_{1}\mathrm{and}40°\mathrm{for}{\mathrm{\theta }}_{2}$ in the above equation.

${F}_{2}={F}_{3}\frac{\mathrm{cos}40°}{\mathrm{cos}60°}\phantom{\rule{0ex}{0ex}}{F}_{2}=1.53{F}_{3}$ …(ii)

Substituting $1.53{\mathrm{F}}_{3}\mathrm{for}{\mathrm{F}}_{2}$in the equation (i):

$0.86{F}_{2}+0.64{F}_{3}=7840N\phantom{\rule{0ex}{0ex}}0.86\left(1.53{F}_{3}\right)+0.64{F}_{3}=7840N\phantom{\rule{0ex}{0ex}}1.95{F}_{3}=7840N\phantom{\rule{0ex}{0ex}}{F}_{3}=4020N$

Substitute $4020\mathrm{N}\mathrm{for}{\mathrm{F}}_{3}$in the equation (ii).

${\mathrm{F}}_{2}=1.5{\mathrm{F}}_{3}\phantom{\rule{0ex}{0ex}}=1.5×4020\mathrm{N}\phantom{\rule{0ex}{0ex}}=6030\mathrm{N}$

Thus, the tension in all of the wires is$7840\mathrm{N},6030\mathrm{N}\mathrm{and}4020\mathrm{N}$ respectively.

## Step 4:- (b) Determination of the strain in each wire

The equation of stress is expressed as:

$\sigma =\frac{F}{A}$

Here, $F$is the force exerted on each wire and $A$is the area of the wire.

The equation of strain is expressed as:

$\epsilon =\frac{∆l}{I}$

Here, $l$is the original length of the wire and $∆l$is the change in the length of the wire.

The equation of the young’s modulus for a wire is expressed as:

$\begin{array}{l}E=\frac{\mathrm{F}/\mathrm{A}}{∆\mathrm{l}/\mathrm{l}}\\ ∆\mathrm{l}/\mathrm{l}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{E}}\end{array}$ …(iii)

Here, $F$is the force exerted on each wire, $A$is the area of the wire, and$∆\mathrm{l}/\mathrm{l}$ is the fractional stretch of the wire.

As the diameter of the wire is same, hence, the area of the wires is calculated as:

$A=\pi {r}^{2}\phantom{\rule{0ex}{0ex}}=\pi {\left(\frac{d}{2}\right)}^{2}$

Here, $r$is the radius of the wire and $d$is the diameter of the wire.

Substitute the value in the above equation:

$A=\pi {\left(\frac{1.1×{10}^{-3}m}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=0.95×{10}^{-6}{m}^{2}$

Now, for the first wire, substituting $0.95×{10}^{-6}{\mathrm{m}}^{2}\mathrm{for}\mathrm{A},7840\mathrm{N}\mathrm{for}\mathrm{F},\mathrm{and}1.3×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}\mathrm{for}\mathrm{E}$in the above equation.

$∆l/l=\frac{7840N/0.95×{10}^{-6}{m}^{2}}{1.3×{10}^{11}N/{m}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{7.448×{10}^{-3}N/{m}^{2}}{1.3×{10}^{11}N/{m}^{2}}\phantom{\rule{0ex}{0ex}}=0.063$

Now, for the second wire, substituting localid="1656912078310" $0.95×{10}^{-6}{\mathrm{m}}^{2}\mathrm{for}\mathrm{A},6030\mathrm{N}\mathrm{for}\mathrm{F}\mathrm{and}1.3×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}\mathrm{for}\mathrm{E}$ in the above equation.

$∆l/l=\frac{6030N/0.95×{10}^{-6}{m}^{2}}{1.3×{10}^{11}N/{m}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{5.7285×{10}^{-3}N/{m}^{2}}{1.3×{10}^{11}N/{m}^{2}}\phantom{\rule{0ex}{0ex}}=0.048$

Now, for the third wire, substituting $0.95×{10}^{-6}{\mathrm{m}}^{2}\mathrm{for}\mathrm{A},4020\mathrm{N}\mathrm{for}\mathrm{F}\mathrm{and}1.3×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}\mathrm{for}\mathrm{E}$ in the above equation.

$∆\mathrm{l}/\mathrm{l}=\frac{4020\mathrm{N}/0.95×{10}^{-6}{\mathrm{m}}^{2}}{1.3×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3.819×{10}^{-3}\mathrm{N}/{\mathrm{m}}^{2}}{1.3×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}}\phantom{\rule{0ex}{0ex}}=0.032$

Thus, the strain in each wire is$0.063,0.048\mathrm{and}0.032$ respectively.