Chapter 10.1, Problem 10.22P

A couple M with a magnitude of 100 N·m is applied as shown to the crank of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the force P required to maintain the equilibrium of the system when (a) θ = 60°, (b) θ = 120°.

Fig. P10.22

To determine

Find the magnitude of the force P required to maintain the equilibrium.

Answer to Problem 10.22P

The magnitude of the force P required is $\underset{\_}{2.05\text{kN}(\to )}$.

Explanation of Solution

Given information:

The magnitude of the couple M is $100\text{N}\cdot \text{m}$.

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 $\theta =60°$.

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 $60°$ for $\theta$ in Equation (1).

$\begin{array}{l}\sin \varphi =\frac{50}{200}\times \sin 60°\\ \varphi =12.504°\end{array}$

Substitute $100\text{N}\cdot \text{m}$ for M, 50 mm for AB, 200 mm for BC, $12.504°$ for $\varphi$, and $60°$ for $\theta$ in Equation (2).

$\begin{array}{l}P\left[50\sin 60°+\left(50\sin 60°\right)\left(\frac{50}{200}\frac{\cos 60°}{\cos 12.504°}\right)\right]-100\text{N}\cdot \text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}=0\\ P\left[43.30127+5.54416\right]-100,000=0\\ P=2.05\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ P=2.05\text{kN}(\to )\end{array}$

Therefore, the magnitude of the force P required is $\underset{\_}{2.05\text{kN}(\to )}$.

To determine

Find the magnitude of the force P required to maintain the equilibrium.

Answer to Problem 10.22P

The magnitude of the force P required is $\underset{\_}{2.65\text{kN}(\to )}$.

Explanation of Solution

Given information:

The magnitude of the couple M is $100\text{N}\cdot \text{m}$.

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 $\theta =120°$.

Calculation:

Refer part (a) for calculation;

Substitute 50 mm for AB, 200 mm for BC, and $120°$ for $\theta$ in Equation (1).

$\begin{array}{l}\sin \varphi =\frac{50}{200}\times \sin 120°\\ \varphi =12.504°\end{array}$

Substitute $100\text{N}\cdot \text{m}$ for M, 50 mm for AB, 200 mm for BC, $12.504°$ for $\varphi$, and $120°$ for $\theta$ in Equation (2).

$\begin{array}{l}P\left[50\sin 120°+\left(50\sin 120°\right)\left(\frac{50}{200}\frac{\cos 120°}{\cos 12.504°}\right)\right]-100\text{N}\cdot \text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}=0\\ P\left[43.30127-5.54416\right]-100,000=0\\ P=2.65\times {10}^3\text{N}\times \frac{\text{1}\text{kN}}{\text{1,000}\text{N}}\\ P=2.65\text{kN}(\to )\end{array}$

Therefore, the magnitude of the force P required is $\underset{\_}{2.65\text{kN}(\to )}$.