Chapter 11, Problem 83PE

To determine

(a)

The Force for the slave cylinder.

Answer to Problem 83PE

  $1.38\times {10}^4\text{N}$.

Explanation of Solution

Given:

Mass of brace, $m_b=15\text{0 kg}$

Mass of load, $m_l=400\text{kg}$

Formula used:

The Force for the slave cylinder is calculated by,

  $F=mg$

Calculation:

Work done is defined in terms of force and distance in angle,

  $W=Fd\sin \theta$

Where W-work done, F-applied force, $d\sin \theta$ -distance in angle.

Force relation in terms of acceleration due to gravity is,

  $F=mg$

Where F-force,m-mass,g-gravity

Work done by brace-shovel and load is given as,

  $\begin{array}{l}{W}_s=W_b+W_l\\ F_s{d}_s\sin {\theta }_s=F_b{d}_b\sin {\theta }_b+F_l{d}_l\sin {\theta }_l\end{array}$

Rewrite the equations.

  $F_s=\frac{F_b{d}_b\sin {\theta }_b+F_l{d}_l\sin {\theta }_l}{d_s\sin {\theta }_s}$

Substitute the $F_b$ and $F_l$ in terms of mass and gravity.

  $F_s=\frac{m_{{}_b}g{d}_b\sin {\theta }_b+m_l{d}_l\sin {\theta }_l}{d_s\sin {\theta }_s}$

Substitute $15\text{0}\text{kg}$ for $m_b$, $400\text{ kg}$ for $m_l$, $1.70\text{ m}$ for $d_l$ 1.10m for $d_b$ ,0.30m for $d_s$.

  $\begin{array}{l}{F}_s=\frac{\left[\left(\text{150 kg}\right)\left(\text{1}\text{.10 m}\right)\text{(sin 30)+(400 kg)(1}\text{.70 m)(sin 30)}\right]}{\text{(0}{\text{.300 m)(sin 90}}^{\text{o}}\text{)}}\text{(9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\text{)}\\ =1.38\times {10}^4\text{N}\end{array}$

Conclusion:

Therefore,the Force for the slave cylinder is $1.38\times {10}^4\text{N}$.

To determine

(b)

The Pressure in the hydraulic fluid.

Answer to Problem 83PE

  $2.81\times {10}^7\text{Pa}$

Explanation of Solution

Given:

Mass of brace, $m_b=15\text{0 kg}$

Mass of load, $m_l=400\text{kg}$

Diameter $=2.5\text{cm}$

Formula used:

Pressure in terms of force and area,

  $P=\frac{F}{A}$

Calculation:

Change the radius from cm to m,

  $\begin{array}{l}r=1.25\text{cm}\\ r=1.25\times {10}^{-2}\text{m}\end{array}$

The area of slave cylinder is given by,

  $A=\pi r^2$

Where r-radius.

Apply the radius value,

  $\begin{array}{l}A=\pi \times {(}^{1.25}\\ A=4.90\times {10}^{-4}{\text{m}}^2\end{array}$

Pressure formula in terms of force and area.

  $P=\frac{F}{A}$

Where p-pressure,F-applied force,A-area perpendicular to the force.

Substitute the value offorce and area,

  $\begin{array}{l}P=\frac{\text{1}{\text{.38×10}}^{\text{4}}\text{N}}{\text{4}{\text{.90×10}}^{\text{-4}}{\text{m}}^{\text{2}}}\\ P=2.81\times {10}^7\text{N}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the obtained Pressure in the hydraulic fluid is $2.81\times {10}^7\text{Pa}$.

To determine

(c)

The force exerted on lever.

Answer to Problem 83PE

  $283\text{N}$.

Explanation of Solution

Given:

Mass of brace, $m_b=15\text{0 kg}$

Mass of load, $m_l=400\text{kg}$

Diameter $=8\text{cm}$

Formula used:

According to Pascal's principle, for hydraulic system,

  $\frac{F_1}{A_1}=\frac{F_2}{A_2}$

The satellite in geosynchronous orbit should possess the orbital period similar to that of earth.

According to Pascal's principle, for hydraulic system,

  $\frac{F_1}{A_1}=\frac{F_2}{A_2}$

Substitute the area formula $\frac{F_1}{\pi r_1}=\frac{F_2}{\pi r_2}$

The equations can be rearranged in terms of,

  $F_1=F_2{\left(\frac{r_1}{r_2}\right)}^2$

Assume the mechanical advantage is acting on master cylinder,

Hence force could be,

  $F_1=\frac{F_2}{5}{\left(\frac{r_1}{r_2}\right)}^2$

Substitute the force value, and the radius of two cylinders,

  $\begin{array}{l}{F}_1=\frac{\text{1}{\text{.38×10}}^{\text{4}}\text{N}}{\text{5}}{\left(\frac{{\text{4×10}}^{\text{3}}\text{m}}{\text{1}{\text{.25×10}}^{\text{-2}}\text{m}}\right)}^{\text{2}}\\ F_1=28\text{3N}\end{array}$

Conclusion:

Therefore,the force exerted on lever is $283\text{N}$.