Chapter 2.2, Problem 2.35P

Knowing that α = 35°, determine the resultant of the three forces shown.

Fig. P2.35

To determine

The resultant of the three forces shown in figure P2.35.

Answer to Problem 2.35P

The resultant of three forces shown in the figure P2.35 is $309\text{N}$ and $86.6°$ below the horizontal towards left.

Explanation of Solution

Refer figure P3.35.

The figure shows three forces of magnitude $100\text{N}$ , $150\text{N}$ , and $200\text{N}$ along with their directions. $\alpha$ is the angle made by the axis of the cylinder and force $100\text{N}$ , $\alpha +30°$ is the angle made by $150\text{N}$ force and the axis and the angle between axis of cylinder and $200\text{N}$ force is $\alpha$.

Since these forces are making an angle with the axis of cylinder, they can be resolved into x and y components.

Write the equation to find the x component of $100\text{N}$ force in the positive x direction.

$F_{x1}=F\cos \alpha$ $\left(\text{I}\right)$

Here, $F_{x1}$ is the x component of $150\text{N}$ force, $F$ is the magnitude of the $150\text{N}$ force, and $\alpha$ is the angle between axis of the cylinder and $100\text{N}$ force.

Write the equation to find the y component of $150\text{N}$ force in the negative y axis.

$F_{y1}=F\sin \alpha$ $\left(\text{II}\right)$

Here, $F_{y1}$ is the y component of $100\text{N}$ force.

Similarly write the equation to find the x component of the $150\text{N}$ force along the positive x direction.

$F_{x2}=F\cos \left(\alpha +30\right)$ $\left(\text{III}\right)$

Here, $F_{x2}$ is the x component of $150\text{N}$ force , $F$ is the magnitude of $150N$ force, and $\alpha +30°$ is the angle between the $150\text{N}$ force vector and the axis of cylinder.

Write the equation to find the y component of $150\text{N}$ force along the negative y direction.

$F_{y2}=F\sin \left(\alpha +30°\right)$ $\left(\text{IV}\right)$

Here, $F_{y2}$ is the y component of $150\text{N}$ force, $F$ is the magnitude of $150\text{N}$ force vector, and $\alpha +30°$ is the angle made by the $150\text{N}$ force vector and the axis of cylinder.

Write the equation to find the x component of the $200\text{N}$ force along the negative x direction.

$F_{x3}=F\cos \alpha$ $\left(\text{V}\right)$

Here, $F_{x3}$ is the x component of the $200\text{N}$ force, $F$ is the magnitude of the $200\text{N}$ force, $\alpha$ is the angle between $200\text{N}$ force vector and the axis of cylinder.

Write the equation to find the y component of $200\text{N}$ force along the negative y direction.

$F_{y3}=F\sin \alpha$ $\left(\text{VI}\right)$

Here, $F_{y3}$ is the y component of $200\text{N}$ force, $F$ is the magnitude of $200\text{N}$ force vector, and $\alpha$ is the angle between the $200\text{N}$ force vector and axis of cylinder.

Express a consolidated form of the x and y components of the three forces as shown in figure 1.

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.

$R_x=F_{x1}+F_{x2}+F_{x3}$ $\left(\text{VII}\right)$

Here, $R_x$ is the resultant of all the three x components of force, $F_{x1}$ is the x component of first force, $F_{x2}$ is the x component of second force, and $F_{x3}$ is the x component of third force.

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

$R_y=F_{y1}+F_{y2}+F_{y3}$ $\left(\text{VIII}\right)$

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

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

Express the vectors $R,{\text{R}}_x,\text{and}{\text{R}}_y$ diagrammatically as shown in figure 2. $\alpha$ is the angle separating the components $R_x$ and the resultant vector $R$.

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

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

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 $\left(\text{X}\right)$ to get $\alpha$.

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

Write the equation to find the y component of the resultant force vector.

$R_y=R\sin \alpha$ $\left(\text{XII}\right)$

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

Rewrite equation $\left(\text{X}\right)$ to get the resultant force vector.

$R=\frac{R_y}{\sin \alpha }$ $\left(\text{XIII}\right)$

Conclusion:

Substitute $100\text{N}$ for $F$, and $35°$ for $\alpha$ in equation $\left(\text{I}\right)$ to get $F_x$.

$\begin{array}{c}{F}_x=\left(100\text{N}\right)\cos 35°\\ =81.915\text{N}\end{array}$

Substitute $-100\text{N}$ for $F$ , and $35°$ for $\alpha$ in equation $\left(\text{II}\right)$ to get $F_y$.

$\begin{array}{c}{F}_y=-\left(100\text{N}\right)\sin 35°\\ =-57.358\text{N}\end{array}$

Substitute $150\text{N}$ for $F$ , and $35°$ for $\alpha$ in equation $\left(\text{III}\right)$  to get $F_x$.

$\begin{array}{c}{F}_x=\left(150\text{N}\right)\cos \left(30+35\right)\\ =\left(150\text{N}\right)\cos \left(65°\right)\\ =63.393\text{N}\end{array}$

Substitute $-150\text{N}$ for $F$ , and $35°$ for $\alpha$ in equation $\left(\text{IV}\right)$ to get $F_y$.

$\begin{array}{c}{F}_y=-\left(150\text{N}\right)\sin 65°\\ =-135.946\text{N}\end{array}$

Substitute $-200\text{N}$ for $F$ , and $35°$ for $\alpha$ in equation $\left(\text{V}\right)$ to get $F_x$.

$\begin{array}{c}{F}_x=-\left(200\text{N}\right)\cos 35°\\ =-163.830\text{N}\end{array}$

Substitute $-200\text{N}$ for $F$ , and $35°$ for $\alpha$ in equation $\left(\text{VI}\right)$ to get $F_y$.

$\begin{array}{c}{F}_y=-\left(200\text{N}\right)\sin 35°\\ =-114.714\text{N}\end{array}$

Substitute $81.915\text{N}$ for $F_{x1}$ , $63.393\text{N}$ for $F_{x2}$ , and $-163.83\text{N}$ for $F_{x3}$ in equation $\left(\text{VII}\right)$ to get $R_x$

$\begin{array}{c}{R}_x=\sqrt{\left(81.915\text{N}\right)+\left(63.393\text{N}\right)+\left(-163.83\text{N}\right)}\\ =-18.522\text{N}\end{array}$

Substitute $-57.358\text{N}$ for $F_{y1}$ , $-135.946\text{N}$ for $F_{y2}$ , and $-114.715\text{N}$ for $F_{y3}$ in equation $\left(\text{VIII}\right)$ to get $R_y$.

$\begin{array}{c}{R}_y=\sqrt{\left(-57.358\text{N}\right)+\left(-135.946\text{N}\right)+\left(-114.714\text{N}\right)}\\ =-308.02\text{N}\end{array}$

Substitute $-18.522\text{N}$ for $R_x$ and $-308.02\text{N}$ for $R_y$ in equation $\left(\text{IX}\right)$ to get $R$.

$R=\left(-18.522\text{N}\right)\widehat{i}+\left(-308.02\text{N}\right)\widehat{j}$

Substitute $308.02\text{N}$ for $R_y$ and $18.522\text{N}$ for $R_x$ in equation $\left(\text{XI}\right)$ to get $\alpha$.

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{308.02\text{N}}{18.522\text{N}}\right)\\ =86.559°\\ \approx 86.6°\end{array}$

Substitute $308.02\text{N}$ for $R_y$ and $86.559°$ for $\alpha$ in equation $\left(\text{XIII}\right)$ to get $R$.

$\begin{array}{c}R=\frac{308.02\text{N}}{\sin \left(86.559°\right)}\\ =309\text{N}\end{array}$

Therefore, the resultant of the three forces shown in figure P2.35 is $\underset{\_}{309\text{N}}$ $86.6°$ below the horizontal towards left.