Chapter 4.2, Problem 4.90P

(a)

To determine

The angle $\theta$ corresponding to the equilibrium condition in the given arrangement.

Answer to Problem 4.90P

The angle $\theta$ corresponding to the equilibrium condition in the given arrangement is $\underset{\_}{59.4°}$.

Explanation of Solution

The free-body diagram corresponding to the arrangement in Fig. P4.89 is given in Figure 1. The reaction at $A$ and $B$ is represented by vector $A$ and vector $B$ respectively. In the equilibrium condition, the reaction $B$ must pass through the point $D$ where $B$ and $W$ intersect.

Write the expression for the distances $AC$ and $AE$.

$AC=L\cos \theta$ (I)

$AE=L\cos \theta$ (II)

Write the expression for the distances $CE$, $BE$, and $BC$ in triangle $EBC$.

$CE=AC+AE$ (III)

$BE=L\sin \theta$ (IV)

$BC=R$ (V)

Write the expression of Pythagoras theorem in the triangle $EBC$.

${\left(CE\right)}^2+{\left(BE\right)}^2={\left(BC\right)}^2$ (VI)

Conclusion:

Use equation (I) and (II) in (III).

$\begin{array}{c}CE=L\cos \theta +L\cos \theta \\ =2L\cos \theta \end{array}$ (VII)

Use equation (IV), (V) and (VII) in (VI).

${\left(2L\cos \theta \right)}^2+{\left(L\sin \theta \right)}^2=R^2$ (VIII)

Simplify equation (VIII).

$\begin{array}{c}4{\cos }^2\theta +{\sin }^2\theta ={\left(\frac{R}{L}\right)}^2\\ 4{\cos }^2\theta +\left(1-{\cos }^2\theta \right)={\left(\frac{R}{L}\right)}^2\\ 3{\cos }^2\theta +1={\left(\frac{R}{L}\right)}^2\\ {\cos }^2\theta =\frac{1}{3}\left[{\left(\frac{R}{L}\right)}^2-1\right]\end{array}$

The expression for the equilibrium condition of the rod is,

${\cos }^2\theta =\frac{1}{3}\left[{\left(\frac{R}{L}\right)}^2-1\right]$ (IX)

Given that the length of the rod is $15\text{in}$, radius of the cylindrical surface is $20\text{in}$, and the weight is $10\text{lb}$.

Substitute $20\text{in}$ for $R$, $15\text{in}$ for $L$ and $10\text{lb}$ for $W$ in equation (IX) and solve for $\theta$.

$\begin{array}{c}{\cos }^2\theta =\frac{1}{3}\left[{\left(\frac{20\text{in}}{15\text{in}}\right)}^2-1\right]\\ \cos \theta =\sqrt{\frac{1}{3}\left[{\left(\frac{20\text{in}}{15\text{in}}\right)}^2-1\right]}\\ \theta ={\cos }^{-1}\left\{\sqrt{\frac{1}{3}\left[{\left(\frac{20\text{in}}{15\text{in}}\right)}^2-1\right]}\right\}\\ =59.39°\end{array}$

Therefore, the angle $\theta$ corresponding to the equilibrium condition in the given arrangement is $\underset{\_}{59.4°}$.

(b)

To determine

The reaction at $A$ and $B$.

Answer to Problem 4.90P

The reaction at $A$ is $\underset{\_}{8.45\text{lb}}$ and that at $B$ is $\underset{\_}{13.09\text{lb}}$.

Explanation of Solution

The free-body diagram corresponding to the arrangement is given in Figure 1.

Write the expression for computing the angle $\alpha$ in the free-body diagram.

$\tan \alpha =\frac{BE}{CE}$ (X)

Use equation (IV) and (VII) in (X).

$\begin{array}{c}\tan \alpha =\frac{L\sin \theta }{2L\cos \theta }\\ =\frac{1}{2}\tan \theta \end{array}$ (XI)

At the equilibrium condition, a force triangle can be drawn using $W$, $A$, and $B$.

Conclusion:

Substitute $59.39°$ for $\theta$ in equation (XI) and solve for $\alpha$.

$\begin{array}{c}\tan \alpha =\frac{1}{2}\tan 59.39°\\ \alpha ={\tan }^{-1}\left(\frac{1}{2}\tan 59.39°\right)\\ =40.2°\end{array}$

The force triangle using $W$, $A$, and $B$ is given in Figure 2. Since $A$ is acting horizontally and $W$ is acting vertically, the force triangle is a right triangle.

Write the trigonometric relation for determining $A$ in the force triangle.

$A=W\tan \alpha$ (XII)

Write the trigonometric relation for det4ermoini ng $B$ in the force triangle.

$B=\frac{W}{\cos \alpha }$ (XIII)

Substitute $40.2°$ for $\alpha$ and $10\text{lb}$ for $W$ in equation (XII) and (XIII) to find the reaction at $A$ and $B$.

$\begin{array}{c}A=\left(10\text{lb}\right)\tan 40.2°\\ =8.45\text{lb}\end{array}$

and,

$\begin{array}{c}B=\frac{10\text{lb}}{\cos 40.2°}\\ =13.09\text{lb}\end{array}$

The reaction at $B$ is acting $49.8°$ clockwise from the horizontal as shown in the force triangle.

Therefore, the reaction at $A$ is $\underset{\_}{8.45\text{lb}}$ and that at $B$ is $\underset{\_}{13.09\text{lb}}$.