Chapter 2.3, Problem 2.51P

To determine

The magnitude of the force exerted on the rod A $\left(F_A\right)$ and B $\left(F_B\right)$.

Answer to Problem 2.51P

The magnitude of the force exerted on the rod A $\left(F_A\right)$ and B $\left(F_B\right)$ are $\underset{\_}{1,303\text{lb}}$ and $\underset{\_}{420\text{lb}}$.

Explanation of Solution

Given information:

The force (P) is 500 lb.

The force (Q) is 650 lb.

The angle $\left(\alpha \right)$ between the force P and Q is $40°$.

The angle $\left(\theta \right)$ between the rod A and B is $50°$.

Calculation:

Draw the free body diagram of the air craft connection as in Figure (1).

Write the vector for force P $\left(P\right)$:

$P=-500j$

Calculate the x component of force $\left(Q_x\right)$ using the relation:

$Q_x=Q\cos \theta$

Substitute 650 lb for $Q$ and $50°$ for $\theta$.

$\begin{array}{c}{Q}_x=650\times \cos \left(50°\right)\\ =-417.8119\end{array}$

Calculate the force $\left(Q_y\right)$ using the relation:

$Q_y=Q\sin \theta$

Substitute 650 lb for Q and $50°$ for $\theta$.

$\begin{array}{c}{Q}_y=-650\times \sin \left(50°\right)\\ =-497.9289\text{lb}\end{array}$

Write the vector $\left(Q\right)$:

$Q=Q_x+Q_y$

Substitute 417.8119 lb for $Q_x$ and 497.9289 lb for $Q_y$.

$Q=417.8119i-497.9289j$

Write the vector for $\left(F_A\right)$:

$F_A=-F_A\cos \left(\theta \right)i+F_A\sin \left(\theta \right)j$

Substitute $50°$ for $\theta$.

$F_A=-F_A\cos \left(50°\right)i+F_A\sin \left(50°\right)j$

Write the vector for $\left(F_B\right)$:

$F_B=+F_{B}i$

Write the expression for the resultant force $\left(R\right)$ using the relation:

$R=P+Q+F_A+F_B$

Substitute $-500j$ for $P$, $417.8119i-497.9289j$ for $Q$, $-F_A\cos \left(50°\right)i+F_A\sin \left(50°\right)j$ for $F_A$, and $F_{B}i$ for $F_B$.

$R=-500j+417.8119i-497.9289j-F_A\cos \left(50°\right)i+F_A\sin \left(50°\right)j+F_{B}i$ (1)

Calculate the magnitude of the force exerted on the rod A $\left(F_A\right)$:

Solve for j component in Equation (1).

$\begin{array}{l}-500-497.9289+F_A\sin \left(50°\right)=0\\ F_A\sin \left(50°\right)=+500+497.9289\\ F_A=\frac{997.9289}{\sin \left(50°\right)}\\ F_A=1,302.70\text{lb}\end{array}$

Calculate the magnitude of the force exerted on the rod B $\left(F_B\right)$:

Solve for i component in Equation (1).

$417.8119-F_A\cos \left(50°\right)+F_B=0$

Substitute 1302.70 lb for $F_A$.

$\begin{array}{l}417.8119-\left(1302.7\right)\cos \left(50°\right)+F_B=0\\ F_B=-417.8119+\left(1302.7\right)\cos \left(50°\right)\\ F_B=419.55\text{lb}\\ F_B\approx \text{420}\text{lb}\end{array}$

Thus, the magnitude of the force exerted on the rod A $\left(F_A\right)$ and B $\left(F_B\right)$ are $\underset{\_}{1,303\text{lb}}$ and $\underset{\_}{420\text{lb}}$.