Chapter 10.1, Problem 10.21P

A 4-kN force P is applied as shown to the piston of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the couple M required to maintain the equilibrium of the system when (a) θ = 30°, (b) θ = 150°.

Fig. P10.21 and P10.22

To determine

Find the couple M required to maintain the equilibrium.

Answer to Problem 10.21P

The magnitude of the couple M is $\underset{\_}{121.8\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)}$.

Explanation of Solution

Given information:

The magnitude of the force P is 4 kN.

The distance between the point A and B is 50 mm.

The distance between the point B and C is 200 mm.

The value of the angle $\theta =30°$.

Calculation:

Show the free-body diagram of the engine system as in Figure 1.

Consider the geometry of the Figure 1.

Use the Law of sines;

$\begin{array}{l}\frac{AB}{\sin \varphi }=\frac{BC}{\sin \theta }\\ \sin \varphi =\frac{AB}{BC}\sin \theta \end{array}$ (1)

Differentiate the equation;

$\begin{array}{l}\cos \varphi \delta \varphi =\frac{AB}{BC}\cos \theta \delta \theta \\ \delta \varphi =\frac{AB}{BC}\frac{\cos \theta }{\cos \varphi }\delta \theta \end{array}$

Find the horizontal displacement $\left(x_C\right)$ at point C using the relation.

$x_C=AB\cos \theta +BC\cos \varphi$

Differentiate the equation;

$\delta x_C=-AB\sin \theta \delta \theta -BC\sin \varphi \delta \varphi$

Substitute $\frac{AB}{BC}\sin \theta$ for $\sin \varphi$ and $\frac{AB}{BC}\frac{\cos \theta }{\cos \varphi }\delta \theta$ for $\delta \varphi$.

$\begin{array}{c}\delta x_C=-AB\sin \theta \delta \theta -BC\left(\frac{AB}{BC}\sin \theta \right)\left(\frac{AB}{BC}\frac{\cos \theta }{\cos \varphi }\delta \theta \right)\\ =-AB\sin \theta \delta \theta -\left(AB\sin \theta \right)\left(\frac{AB}{BC}\frac{\cos \theta }{\cos \varphi }\delta \theta \right)\end{array}$

Use the principle of virtual work;

$\begin{array}{l}\delta U=0\\ -P\delta x_C-M\delta \theta =0\end{array}$

Substitute $\left[-AB\sin \theta \delta \theta -\left(AB\sin \theta \right)\left(\frac{AB}{BC}\frac{\cos \theta }{\cos \varphi }\delta \theta \right)\right]$ for $\delta x_C$.

$\begin{array}{l}-P\left[-AB\sin \theta \delta \theta -\left(AB\sin \theta \right)\left(\frac{AB}{BC}\frac{\cos \theta }{\cos \varphi }\delta \theta \right)\right]-M\delta \theta =0\\ P\left[AB\sin \theta +\left(AB\sin \theta \right)\left(\frac{AB}{BC}\frac{\cos \theta }{\cos \varphi }\right)\right]-M=0\end{array}$ (2)

Substitute 50 mm for AB, 200 mm for BC, and $30°$ for $\theta$ in Equation (1).

$\begin{array}{l}\sin \varphi =\frac{50}{200}\times \sin 30°\\ \varphi =7.181°\end{array}$

Substitute 4 kN for P, 50 mm for AB, 200 mm for BC, $7.181°$ for $\varphi$, and $30°$ for $\theta$ in Equation (2).

$\begin{array}{l}4\left[50\sin 30°+\left(50\sin 30°\right)\left(\frac{50}{200}\frac{\cos 30°}{\cos 7.181°}\right)\right]-M=0\\ 4\left[25+25\left(0.21822\right)\right]-M=0\\ M=121.8\text{kN}\cdot \text{mm}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\\ M=121.8\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)\end{array}$

Therefore, the magnitude of the couple M is $\underset{\_}{121.8\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)}$.

To determine

Find the couple M required to maintain the equilibrium.

Answer to Problem 10.21P

The magnitude of the couple M is $\underset{\_}{78.2\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)}$.

Explanation of Solution

Given information:

The magnitude of the force P is 4 kN.

The distance between the point A and B is 50 mm.

The distance between the point B and C is 200 mm.

The value of the angle $\theta =150°$.

Calculation:

Refer part (a) for calculation;

Substitute 50 mm for AB, 200 mm for BC, and $150°$ for $\theta$ in Equation (1).

$\begin{array}{l}\sin \varphi =\frac{50}{200}\times \sin 150°\\ \varphi =7.181°\end{array}$

Substitute 4 kN for P, 50 mm for AB, 200 mm for BC, $7.181°$ for $\varphi$, and $150°$ for $\theta$ in Equation (2).

$\begin{array}{l}4\left[50\sin 150°+\left(50\sin 150°\right)\left(\frac{50}{200}\frac{\cos 150°}{\cos 7.181°}\right)\right]-M=0\\ 4\left[25-25\left(0.21822\right)\right]-M=0\\ M=78.2\text{kN}\cdot \text{mm}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\\ M=78.2\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)\end{array}$

Therefore, the magnitude of the couple M is $\underset{\_}{78.2\text{N}\cdot \text{m}\left(\text{Counter}\text{clockwise}\right)}$.