Chapter 2.2, Problem 2.22P

Determine the x and y components of each of die forces shown.

Fig. P2.22

To determine

To determine the x and y components of the forces shown in figure P2.22.

Answer to Problem 2.22P

The x component of the $800\text{N}$ force is $\underset{\_}{+640\text{N}}$ and the y component of $800\text{N}$ force is $\underset{\_}{+480\text{N}}$.

The x component of the $424\text{N}$ force is $\underset{\_}{-224\text{N}}$ and the y component of $424\text{N}$ force is $\underset{\_}{-360\text{N}}$.

The x component of the $408\text{N}$ force is $\underset{\_}{+192\text{N}}$ and the y component of $408\text{N}$ force is $\underset{\_}{-360\text{N}}$

Explanation of Solution

Refer figure P2.22.

The simplified diagram of the figure P2.22 is shown in figure 1 below. Figure 1 shows all the three components of force and the corresponding position vectors and their magnitudes. Position vectors $OA,OB\text{and}OC$ represents the resultant force components along X, Y and Z axis.

Let $OA$ be the position vector corresponding to $800\text{N}$ force, let $OB$ be the position vector corresponding to $424\text{N}$ and let $OC$ be the position vector corresponding to $408\text{N}$.

Write the equation to find the magnitude of position vector $OA$.

$\left|OA\right|=\sqrt{x^2+y^2}$                                                                                                      …...$\left(\text{I}\right)$

Here, $\left|OA\right|$ is the magnitude of position vector $OA$, $x$ is the x component of position vector $OA$, and $y$ is the y component of position vector $OA$.

Write the equation to find the magnitude of position vector $OB$.

$\left|OB\right|=\sqrt{x^2+y^2}$ $\left(\text{II}\right)$

Here, $\left|OB\right|$ is the magnitude of position vector $OB$, $x$ is the x component of position vector $OB$, and $y$ is the y component of position vector $OB$.

Write the equation to find the magnitude of position vector $OC$.

$\left|OC\right|=\sqrt{x^2+y^2}$ $\left(\text{III}\right)$

Here, $\left|OC\right|$ is the magnitude of position vector $OC$, $x$ is the x component of position vector $OC$, and $y$ is the y component of position vector $OC$.

Let $F$ be the resultant force acting on the body. The force vector can be resolved into x and y components by taking the cos and sin of the resultant force.

Write the equation to find the x component of the $800\text{N}$ force.

$F_x=F_{800}\cos a$ $\left(\text{IV}\right)$

Here, $F_x$ is the $x$ component of the $800\text{N}$ force, and $F_{800}$ is the magnitude of the resultant force.

$\cos a$ is the cosine of the x and y components of $800\text{N}$ force which is equal to $\frac{adjacentside}{hypotenuse}$

$F_x=F_{800}\cos \left[\frac{adjacentside}{hypotenuse}\right]$ $\left(\text{V}\right)$

Here, $hypotenuse$ is otherwise the resultant force.

Similarly write the equation to find the y component of $800\text{N}$ force.

$F_y=F_{800}\text{sin}\text{a}$ $\left(\text{VI}\right)$

Here, $F_y$ is the $y$ component of the $800\text{N}$ force, $F_{800}$ is the resultant force, and $\sin a$ is the sine of the x and y components of position vectors of $800\text{N}$ force which is equal to $\frac{adjacentside}{hypotenuse}$

Rewrite the equation of y component of $800\text{N}$ force.

$F_y=F_{800}\sin \left[\frac{adjacentside}{hypotenuse}\right]$ $\left(\text{VII}\right)$

Here, $hypotenuse$ is otherwise the resultant force.

Write the equation to find the x component of the $424\text{N}$ force.

$F_x=F_{424}\cos a$ $\left(\text{VIII}\right)$

Here, $F_x$ is the $x$ component of the $424\text{N}$ force, $F_{424}$ is the magnitude of resultant force, and

$\cos a$ is the cosine of the x and y components of $424\text{N}$ force which is equal to $\frac{adjacentside}{hypotenuse}$.

$F_x=F_{424}\cos \left[\frac{adjacentside}{hypotenuse}\right]$ $\left(\text{IX}\right)$

Here, $hypotenuse$ is otherwise the resultant force.

Similarly write the equation to find the y component of $424\text{N}$ force.

$F_y=F_{424}\text{sin}\text{a}$ $\left(\text{X}\right)$

Here, $F_y$ is the $y$ component of the $424\text{N}$ force, $F_{424}$ is the magnitude of resultant force, and $\sin a$ is the sine of the x and y components of position vectors of $424\text{N}$ force which is equal to $\frac{adjacentside}{hypotenuse}$.

$F_y=F_{424}\sin \left[\frac{adjacentside}{hypotenuse}\right]$ $\left(\text{XI}\right)$

Here, $hypotenuse$ is otherwise the resultant force.

Write the equation to find the x component of the $408\text{N}$ force.

$F_x=F_{408}\cos a$ $\left(\text{XII}\right)$

Here, $F_x$ is the $x$ component of the $408\text{N}$ force, $F_{408}$ is the magnitude of resultant force, and

$\cos a$ is the cosine of the x and y components of $408\text{N}$ force which is equal to $\frac{adjacentside}{hypotenuse}$

$F_x=F_{408}\cos \left[\frac{adjacentside}{hypotenuse}\right]$ $\left(\text{XIII}\right)$

Here, $hypotenuse$ is otherwise the resultant force.

Similarly write the equation to find the y component of $408\text{N}$ force.

$F_y=F_{408}\text{sin}\text{a}$

Here, $F_y$ is the $y$ component of the $408\text{N}$ force, $F_{408}$ is the magnitude of resultant force, and $\sin a$ is the sine of the x and y components of position vectors of $408\text{N}$ force which is equal to $\frac{adjacentside}{hypotenuse}$.

$F_y=F_{408}\sin \left[\frac{adjacentside}{hypotenuse}\right]$ $\left(\text{XIV}\right)$

Here, $hypotenuse$ is otherwise the resultant magnitude of the force.

Conclusion:

Substitute $800\text{mm}$ for $x$ and $600\text{mm}$ for $y$ in the equation $\left(\text{I}\right)$ to get $\left|OA\right|$.

$\begin{array}{c}\left|OA\right|=\sqrt{{\left(800\text{mm}\right)}^2+{\left(600\text{mm}\right)}^2}\\ =1000\text{mm}\end{array}$

Substitute $560\text{mm}$ for $x$ and $900\text{mm}$ for $y$ in the equation $\left(\text{II}\right)$ to get $\left|OB\right|$.

$\begin{array}{c}\left|OB\right|=\sqrt{{\left(900\text{mm}\right)}^2+{\left(560\text{mm}\right)}^2}\\ =1060\text{mm}\end{array}$

Substitute $480\text{mm}$ for $x$ and $900\text{mm}$ for $y$ in the equation $\left(\text{III}\right)$ to get $\left|OC\right|$.

$\begin{array}{c}\left|OA\right|=\sqrt{{\left(900\text{mm}\right)}^2+{\left(480\text{mm}\right)}^2}\\ =1020\text{mm}\end{array}$

Substitute $800\text{N}$ for $F_x$ and $800\text{mm}$ for $adjacentside$, and $1000\text{mm}$ for $hypotenuse$ in the equation $\left(\text{V}\right)$ to get $F_x$.

$\begin{array}{c}{F}_x=800\text{N}\text{cos}\left[\frac{800\text{mm}}{1000\text{mm}}\right]\\ =+640\text{N}\end{array}$

Substitute $800\text{N}$ for $F_y$ and $600\text{mm}$ for $adjacentside$, and $1000\text{mm}$ for $hypotenuse$ in the equation $\left(\text{VII}\right)$ to get $F_y$.

$\begin{array}{c}{F}_x=800\text{N}\text{cos}\left[\frac{600\text{mm}}{1000\text{mm}}\right]\\ =+480\text{N}\end{array}$

Substitute $-424\text{N}$ for $F_{424}$ and $560\text{mm}$ for $adjacentside$, and $1060\text{mm}$ for $hypotenuse$ in the equation $\left(\text{IX}\right)$ to get $F_x$.

$\begin{array}{c}{F}_x=\left(-424\text{N}\right)\text{cos}\left[\frac{560\text{mm}}{1060\text{mm}}\right]\\ =-224\text{N}\end{array}$

Substitute $424\text{N}$ for $F_y$ and $900\text{mm}$ for $adjacentside$, and $1060\text{mm}$ for $hypotenuse$ in the equation $\left(\text{XI}\right)$ to get $F_y$.

$\begin{array}{c}{F}_x=424\text{N}\text{cos}\left[\frac{900\text{mm}}{1060\text{mm}}\right]\\ =-360\text{N}\end{array}$

Substitute $-408\text{N}$ for $F_{408}$ and $480\text{mm}$ for $adjacentside$, and $1020\text{mm}$ for $hypotenuse$ in the equation $\left(\text{XIII}\right)$ to get $F_x$.

$\begin{array}{c}{F}_x=\left(408\text{N}\right)\text{cos}\left[\frac{480\text{mm}}{1020\text{mm}}\right]\\ =192\text{N}\end{array}$

Substitute $408\text{N}$ for $F_y$ and $900\text{mm}$ for $adjacentside$, and $1020\text{mm}$ for $hypotenuse$ in the equation $\left(\text{XIV}\right)$ to get $F_y$.

$\begin{array}{c}{F}_y=\left(-408\text{N}\right)\text{cos}\left[\frac{900\text{mm}}{1020\text{mm}}\right]\\ =-360\text{N}\end{array}$

Therefore, The x component of the $800\text{N}$ force is $\underset{\_}{+640\text{N}}$ and the y component of $800\text{N}$ force is $\underset{\_}{+480\text{N}}$.

The x component of the $424\text{N}$ force is $\underset{\_}{-224\text{N}}$ and the y component of $424\text{N}$ force is $\underset{\_}{-360\text{N}}$.

The x component of the $408\text{N}$ force is $\underset{\_}{+192\text{N}}$ and the y component of $408\text{N}$ force is $\underset{\_}{-360\text{N}}$