Chapter 11, Problem 1FP

Determine the required magnitude of force P to maintain equilibrium of the linkage at θ = 60°. Each link has a mass of 20 kg.

To determine

The required magnitude of force P to maintain equilibrium.

Answer to Problem 1FP

The required magnitude of force P to maintain equilibrium is $\underset{\_}{56.6\text{N}}$.

Explanation of Solution

Given information:

The mass of each link $\left(m\right)$ is 20 kg.

The given angle $\theta$ is $60°$.

Show the free body diagram of the forces acting on the link as in Figure (1).

Observation:

The weight W and the horizontal component of P do positive work, when the position coordinates $\left(x_C\text{and}{y}_G\right)$ undergo a positive virtual displacement.

Virtual displacement:

Write the position coordinate about x axis at point C.

$x_C=\left(l_{AB}+l_{BC}\right)\cos \theta$ (I)

Here, the length of the link AB is $l_{AB}$ and the length of the link BC is $l_{BC}$.

Write the position coordinate about y axis about the center of the link.

$y_G=\frac{l_{AB}}{2}\sin \theta$ (II)

Determine the weight of the link W.

$W=mg$ (III)

Here, the acceleration due to gravity is g.

Virtual work equation:

Write the virtual work equation to determine the magnitude of force P.

$\begin{array}{l}\delta U=0\\ 2W\delta y_G+P\delta x_C=0\end{array}$ (IV)

Here, the angle of force $F_{AC}$ is $\theta$.

Conclusion:

Differentiate Equation (I) with respect to $\theta$.

$\begin{array}{l}\frac{\delta x_C}{\delta \theta }=\left(l_{AB}+l_{BC}\right)\cos \theta \\ \delta x_C=-\left(l_{AB}+l_{BC}\right)\sin \theta \delta \theta \end{array}$ (V)

Substitute 1.5 m for $l_{AB}$ and 1.5 m for $l_{BC}$ in Equation (V).

$\begin{array}{c}\delta x_C=-\left(1.5+1.5\right)\sin \theta \delta \theta \\ =-3\sin \theta \delta \theta \end{array}$

Differentiate Equation (II) with respect to $\theta$.

$\begin{array}{l}\frac{\delta y_G}{\delta \theta }=\frac{l_{AB}}{2}\sin \theta \\ \delta y_G=\frac{l_{AB}}{2}\cos \theta \delta \theta \end{array}$

Substitute 1.5 m for $l_{AB}$.

$\begin{array}{c}\delta y_G=\frac{1.5}{2}\cos \theta \delta \theta \\ =0.75\cos \theta \delta \theta \end{array}$

Substitute 20 kg for m and $9.81\text{m}/{\text{s}}^{\text{2}}$ for g in Equation (III).

$\begin{array}{c}W=20\times 9.81\\ =196.2\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\times \frac{1\text{N}}{1\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}}\\ =196.2\text{N}\end{array}$

Substitute 196.2 N for W, $0.75\cos \theta \delta \theta$ for $\delta y_G$, $60°$ for $\theta$, and $\left(-3\sin \theta \delta \theta \right)$ for $\delta x_C$ in Equation (IV).

$\begin{array}{l}2\left(196.2\right)\left(0.75\cos \theta \delta \theta \right)+P\left(-3\sin \theta \delta \theta \right)=0\\ \delta \theta \left[\left(294.3\cos 60°\right)-3P\sin 60°\right]=0\end{array}$

Here, the value of $\delta \theta \ne 0$.

Modify the Equation,

$\begin{array}{l}\left(294.3\cos 60°\right)-3P\sin 60°=0\\ 147.15=2.598P\\ \frac{147.15}{2.598}=P\\ P=56.6\text{N}\end{array}$

Thus, the required magnitude of force P to maintain equilibrium is $\underset{\_}{56.6\text{N}}$.