Chapter 4, Problem 25P

To determine

The magnitude of $F_B$ and ${\overrightarrow{F}}_A+{\overrightarrow{F}}_B$ .

Answer to Problem 25P

Solution:

$F_B=6310.7\text{N}$ and ${\overrightarrow{F}}_A+{\overrightarrow{F}}_B=6355.25\text{N}$ .

Explanation of Solution

From the free body diagram below, the component of force $F_B$ along x -direction balances the component of $F_A$ along x -direction.

Given:

  $F_A=4500\text{N}$

Calculation:

Using equilibrium of force in the x -direction,

  $\begin{array}{l}{F}_A\text{sin}{\text{48}}^{\text{o}}=F_B\text{sin}{\text{32}}^{\text{o}}\end{array}$

$\begin{array}{c}(4500)\text{sin}{\text{48}}^{\text{o}}=F_B\text{sin}{\text{32}}^{\text{o}}\end{array}$

     $\begin{array}{c}{F}_B=6310.7\text{N}\end{array}$

From the free body diagram, the component of force $F_B$ along x -direction balances the component of $F_A$ along x -direction. Hence, the net force is along the y -direction by adding the component of each force along the y-direction

The net force is given

$\begin{array}{c}{\overrightarrow{F}}_A+{\overrightarrow{F}}_B=F_A{\text{cos 48}}^{\text{ο}}+F_B\text{sin}{\text{32}}^{\text{o}}\end{array}$

     $\begin{array}{c}=(4500)\text{cos}{\text{48}}^{\text{ο}}+(6310.7)\sin {\text{32}}^{\text{o}}\end{array}$

     $\begin{array}{c}=6355.25\text{N}\end{array}$

Conclusion:

Thus, $F_B=6310.7\text{N}$ and ${\overrightarrow{F}}_A+{\overrightarrow{F}}_B=6355.25\text{N}$ .