Chapter 3, Problem 1P

3–1* to 3–4 Sketch a free-body diagram of each element in the figure. Compute the magnitude and direction of each force using an algebraic or vector method, as specified.

To determine

The free body diagram of each element of the given figure.

The magnitude and direction of each force.

Answer to Problem 1P

The free body diagram of element OB is shown in Figure (1).

The free body diagram of element BC is shown in Figure (2).

The free body diagram of pin joint C is shown in Figure (3).

The free body diagram of pin joint O is shown in Figure (4).

The magnitude of force at O is $\underset{\_}{\text{66}\text{.66}\text{lbf}}$ in the positive $y$ direction.

The magnitude of force at B is $\underset{\_}{\text{33}\text{.33}\text{lbf}}$ in the positive $y$ direction.

The magnitude of force at C is $\underset{\_}{\text{33}\text{.33}\text{lbf}}$ in the negative $y$ direction.

Explanation of Solution

Figure (1) shows the free body diagram of the element OB.

Figure (1)

Write the net moment at point O on the link OB.

    $F_{\text{BV}}\times l_1-W\times l_2=0$                                                        (I)

Here, perpendicular distance between point O and $F_{\text{BV}}$ is $l_1$, perpendicular distance between point O and the applied force is $l_2$, force acting at point A is $W$.

Write the net reaction force in the vertical direction on the link OB.

    $F_{\text{OV}}+F_{\text{BV}}-W=0$                                                        (II)

Here, force acting at point O is $F_{\text{OV}}$, force acting at point B is $F_{\text{BV}}$.

Write the net reaction force in the horizontal direction on the link OB.

    $F_{\text{OH}}-F_{\text{BH}}=0$                                                             (III)

Here, force acting at point O is $F_{\text{OH}}$, force acting at point B is $F_{\text{BH}}$.

Figure (2) shows the free body diagram of element BC.

Figure (2)

Write the net reaction force in the horizontal direction on the link BC.

    $F_{\text{BH}}-F_{\text{CH}}=0$                                               (IV)

Here, force acting at point B is $F_{\text{BH}}$, force acting at point C is $F_{\text{CH}}$.

Write the net reaction force in the vertical direction on the link OB.

    $F_{\text{CV}}-F_{\text{BV}}=0$                                                (V)

Here, force acting at point C is $F_{\text{CV}}$, force acting at point B is $F_{\text{BV}}$.

Figure (3) shows the free body diagram of pin joint C.

Figure (3)

Write the expression of net reaction force at pin joint C in horizontal direction.

    $F_{\text{CH}}=0$                                               (VI)

Here, force acting at point C in the horizontal direction is $F_{\text{CH}}$.

The force acting in the vertical direction at point C is balanced by the reaction force by link BC at point C.

Figure (4) shows the free body diagram of the pivot point O.

Figure (4)

Write the expression of net reaction force at pin joint O in horizontal direction.

    $F_{\text{OH}}=0$                                               (VII)

Here, force acting at point O in the horizontal direction is $F_{\text{OH}}$.

The force acting in the vertical direction at point O is balanced by the reaction force by link OB at point O.

Conclusion

Substitute $100\text{lbf}$ for $W$ in Equation (I).

    $\begin{array}{l}{F}_{BV}\times \left(18\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)-\left(100\text{lbf}\times 6\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)=0\\ F_{\text{BV}}\times \frac{18\text{ft}}{12}=\frac{600\text{lbf}\cdot \text{ft}}{12}\\ F_{\text{BV}}=\frac{600}{18}\text{lbf}\\ F_{\text{BV}}=\text{33}\text{.33}\text{lbf}\end{array}$

Thus, the vertical component of the force at point $B$ is $33.33\text{lbf}$.

Substitute $\text{33}\text{.33}\text{lbf}$ for $F_{\text{BV}}$ and $100\text{}\text{lbf}$ for $W$ in Equation (II).

    $\begin{array}{l}{F}_{\text{OV}}+33.33\text{lbf}-100\text{lbf}=0\\ F_{\text{OV}}=\left(100-33.33\right)\text{lbf}\\ F_{\text{OV}}=\text{66}\text{.66}\text{lbf}\end{array}$

Thus, the vertical component of the force at point $O$ is $66.66\text{lbf}$.

Substitute $33.33\text{lbf}$ for $F_{\text{BV}}$ in Equation (V).

    $\begin{array}{l}{F}_{\text{CV}}-33.33\text{lbf}=0\\ F_{\text{CV}}=33.33\text{lbf}\end{array}$

Thus, the vertical component of the force at point $C$ is $33.33\text{lbf}$.

Substitute $0$ for $F_{\text{CH}}$ in Equation (IV).

    $\begin{array}{l}{F}_{\text{BH}}-0=0\\ F_{\text{BH}}=0\end{array}$

Thus, the horizontal component of the force at point $B$ is $0$.

Substitute $0$ for $F_{\text{BH}}$ in Equation (III).

    $\begin{array}{l}{F}_{\text{OH}}-0=0\\ F_{\text{OH}}=0\end{array}$

Thus, the horizontal component of the force at point $O$ is $0$.

Thus, the magnitude of force at O is $\underset{\_}{\text{66}\text{.66}\text{lbf}}$ in the positive $y$ direction.

Thus, the magnitude of force at B is $\underset{\_}{\text{33}\text{.33}\text{lbf}}$ in the positive $y$ direction.

Thus, the magnitude of force at C is $\underset{\_}{\text{33}\text{.33}\text{lbf}}$ in the negative $y$ direction.