Chapter 2.2, Problem 2.36P

Knowing that the tension in cable BC is 725 N, determine the resultant of the three forces exerted at point B of beam AB.

Fig. P2.36

To determine

The resultant of the three forces shown in figure P2.36.

Answer to Problem 2.36P

The resultant of three forces shown in the figure P2.36 is $\underset{\_}{226\text{N}}$.

Explanation of Solution

Figure 1 below shows the three forces acting on point A and their directions.

The figure shows three forces of magnitude $725\text{N}$ , $500\text{N}$ , and $780\text{N}$ along with their directions. Angle $a$ is the angle between the beam AB and the wire BC.

Since these forces make an angle with the axis, they can be resolved into x and y components.

The tension in the wire BC is $725\text{N}$ which has a length of $1160\text{mm}$. All the three forces exert its resultant force at point marked as point B in figure P2.36.

Write the equation to find the x component of $725\text{N}$ tension in the wire BC.

$F_{x1}=F_1\cos a$ $\left(\text{I}\right)$

Here, $F_{x1}$ is the x component of $725\text{N}$ tension, $F_1$ is $\text{tension}$ due to $725\text{N}$ , and $a$ is the angle between beam and direction of tension.

$\cos$ is the ratio of the adjacent side and the hypotenuse of a triangle considered in the figure.

Substitute $\frac{adjacentside}{hypotenuse}$ for $\cos a$ in equation $\left(\text{I}\right)$ to get $F_{x1}$.

$F_{x1}=F_1\left(\frac{adjacentside}{hypotenuse}\right)$ $\left(\text{II}\right)$

Write the equation to find the y component of the tension in the wire.

$F_y{}_1=F_1\sin a$ $\left(\text{III}\right)$

Here, $F_y{}_1$ is the y component of tension due to $F_1$.

$\sin a$ is the ratio of opposite side and the hypotenuse of the triangle considered.

Substitute $\frac{oppositeside}{hypotenuse}$ for $\sin a$ in equation $\left(\text{III}\right)$ to get $F_{y1}$.

$F_{y1}=F_1\left(\frac{oppositeside}{hypotenuse}\right)$ $\left(\text{IV}\right)$

Similarly write the equation to find the x component of the $500\text{N}$ force along the negative x direction.

$F_{x2}=F_2\cos b$ $\left(\text{V}\right)$

Here, $F_{x2}$ is the x component of $500\text{N}$ force , $F_2$ is the magnitude of $500N$ force, and $b$ is the angle between the $500\text{N}$ force vector and the beam.

$\cos$ is the ratio of the adjacent side and the hypotenuse of a triangle considered in the figure.

Substitute $\frac{adjacentside}{hypotenuse}$ for $\cos b$ in equation $\left(\text{V}\right)$ to get $F_{x2}$.

$F_{x2}=F_2\cos \left(\frac{adjacentside}{hypotenuse}\right)$ $\left(\text{VI}\right)$

Write the equation to find the y component of $\text{500}\text{N}$ force along the negative y direction.

$F_{y2}=F_2\sin b$ $\left(\text{VII}\right)$

Here, $F_{y2}$ is the y component of $500\text{N}$ force.

$\sin b$ is the ratio of opposite side and the hypotenuse of the triangle considered.

Substitute $\frac{oppositeside}{hypotenuse}$ for $\sin b$ in equation $\left(\text{VII}\right)$ to get $F_{y2}$.

$F_{y2}=F_2\left(\frac{oppositeside}{hypotenuse}\right)$ $\left(\text{VIII}\right)$

Write the equation to find the x component of the $780\text{N}$ force along the positive x direction.

$F_{x3}=F_3\cos c$ $\left(\text{IX}\right)$

Here, $F_{x3}$ is the x component of the $780\text{N}$ force, $F_3$ is the magnitude of the $780\text{N}$ force, $c$ is the angle between $\text{780}\text{N}$ force vector and the beam.

$\cos$ is the ratio of the adjacent side and the hypotenuse of a triangle considered in the figure.

Substitute $\frac{adjacentside}{hypotenuse}$ for $\cos c$ in equation $\left(\text{IX}\right)$ to get $F_{x3}$.

$F_{x3}=F_3\left(\frac{adjacentside}{hypotenuse}\right)$ $\left(\text{X}\right)$

Write the equation to find the y component of $780\text{N}$ force along the negative y direction.

$F_{y3}=F_3\sin c$ $\left(\text{XI}\right)$

Here, $F_{y3}$ is the y component of $780\text{N}$ force.

$\sin c$ is the ratio of opposite side and the hypotenuse of the triangle considered.

Substitute $\frac{oppositeside}{hypotenuse}$ for $c$ in equation $\left(\text{XI}\right)$ to get $F_y$.

$F_{y3}=F_3\left(\frac{oppositeside}{hypotenuse}\right)$ $\left(\text{XII}\right)$

Let $R_x$ and $R_y$ be the x and y component of the resultant forces and $R$ be the resultant force vector.

Write the equation to find the resultant of all the three x components.

$\begin{array}{c}{R}_x=F_{x1}+F_{x2}+F_{x3}\\ =\sum F_x\end{array}$ $\left(\text{XIII}\right)$

Here, $R_x$ is the resultant of all the three x components of force, and $\sum F_x$ is the sum of all the three x components of forces.

Write the equation to find the resultant of all the three y components.

$\begin{array}{c}{R}_y=F_{y1}+F_{y2}+F_{y3}\\ =\sum F_y\end{array}$ $\left(\text{XIV}\right)$

Here, $\sum F_y$ is the sum of all the y components of three forces acting on point B.

Write the general expression of a vector with $R_x$ and $R_y$ as components.

$R=R_x\widehat{i}+R_y\widehat{j}$ $\left(\text{XV}\right)$

Express the vectors $R,{\text{R}}_x,\text{and}{\text{R}}_y$ diagrammatically as shown in figure 1. $\alpha$ is the angle separating the components $R_x$ and the resultant vector $R$.

Write the equation to find the resultant of vectors $R_x$ and $R_y$.

$R=\sqrt{R_x{}^2+R_y{}^2}$ $\left(\text{XVI}\right)$

Here, $R$ is the resultant force, $R_x$ is the resultant x component of force, and $R_y$ is the resultant of y component of force.

Write the equation to find the tangent of angle $\alpha$.

$\tan \alpha =\frac{R_y}{R_x}$ $\left(\text{XVII}\right)$

Here, $\alpha$ is the angle between component $R_x$, and $R$ , $R_y$ is the y component of the resultant force, and $R_x$ is the x component of the resultant force.

Rewrite equation $\left(\text{XII}\right)$ to get $\alpha$.

$\alpha ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$ $\left(\text{XVIII}\right)$

Conclusion:

Substitute $-725\text{N}$ for $F_1$, $840\text{mm}$ for $adjacentside$  and $1160\text{mm}$ for $hypotenuse$ in equation $\left(\text{II}\right)$ to get $F_{x1}$.

$\begin{array}{c}{F}_{x1}=-\left(725\text{N}\right)\frac{840\text{mm}}{1160\text{mm}}\\ =-525\text{N}\end{array}$

Substitute $725\text{N}$ for $F_1$ ,$800\text{mm}$ for $oppositeside$  and $1160\text{mm}$ for $hypotenuse$ in equation $\left(\text{IV}\right)$ to get $F_{y1}$.

$\begin{array}{c}{F}_{y1}=\left(725\text{N}\right)\frac{800\text{mm}}{1160\text{mm}}\\ =500\text{N}\end{array}$

Substitute $-500\text{N}$ for $F_2$ , $3\text{mm}$ for $adjacentside$ and $5\text{mm}$ for $hypotenuse$ in equation $\left(\text{VI}\right)$  to get $F_{x2}$.

$\begin{array}{c}{F}_{x2}=-\left(500\text{N}\right)\frac{3\text{mm}}{5\text{mm}}\\ =-300\text{N}\end{array}$

Substitute $-500\text{N}$ for $F_2$ , $5\text{mm}$ for $hypotenuse$ and $4\text{mm}$ for $oppositeside$ in equation $\left(\text{VIII}\right)$ to get $F_{y2}$.

$\begin{array}{c}{F}_{y2}=-\left(500\text{N}\right)\frac{4\text{mm}}{5\text{mm}}\\ =-400\text{N}\end{array}$

Substitute $780\text{N}$ for $F_3$ , $12\text{mm}$ for $adjacentside$ and $13\text{mm}$ for $hypotenuse$ in equation $\left(\text{X}\right)$ to get $F_{x3}$.

$\begin{array}{c}{F}_{x3}=\left(780\text{N}\right)\frac{12\text{mm}}{13\text{mm}}\\ =720\text{N}\end{array}$

Substitute $-780\text{N}$ for $F_3$ , $5\text{mm}$ for $oppositeside$ and $13\text{mm}$ for $hypotenuse$ in equation $\left(\text{XII}\right)$ to get $F_{y3}$.

$\begin{array}{c}{F}_{y3}=-\left(780\text{N}\right)\frac{5\text{mm}}{13\text{mm}}\\ =-300\text{N}\end{array}$

Substitute $-525\text{N}$, $-300\text{N}$ , and $720\text{N}$ for $\sum F_x$ in equation $\left(\text{XIII}\right)$ to get $R_x$

$\begin{array}{c}{R}_x=-525\text{N-300}\text{N+720}\text{N}\\ =-105\text{N}\end{array}$

Substitute $500\text{N}$ , $-400\text{N}$ , and $-300\text{N}$ for $\sum F_y$ in equation $\left(\text{XIV}\right)$ to get $R_y$.

$\begin{array}{c}{R}_y=500\text{N}-400\text{N}-300\text{N}\\ =-200\text{N}\end{array}$

Substitute $-105\text{N}$ for $R_x$ and $-200\text{N}$ for $R_y$ in equation $\left(\text{XV}\right)$ to get $R$.

$R=\left(-105\text{N}\right)\widehat{i}+\left(-200\text{N}\right)\widehat{j}$

Substitute $-105\text{N}$ for $R_x$ and $-200\text{N}$ for $R_y$ in equation $\left(\text{XVI}\right)$ to get $R$.

$\begin{array}{c}R=\sqrt{{\left(-105\text{N}\right)}^2+{\left(-200\text{N}\right)}^2}\\ =225.89\text{N}\end{array}$

Substitute $200\text{N}$ for $R_y$ and $105\text{N}$ for $R_x$ in equation $\left(\text{XIII}\right)$ to get $\alpha$.

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{200\text{N}}{105\text{N}}\right)\\ =62.3°\end{array}$

Therefore, the resultant of the three forces acting at point B shown in figure P2.36 is $\underset{\_}{226\text{N}}$.