91影视

The given data can be listed below,

• the magnitude of force vector$\stackrel{鈬赌}{\mathrm{A}}\mathrm{is},100\mathrm{N}.$
• the magnitude of force vector$\stackrel{鈬赌}{\mathrm{B}}\mathrm{is},80\mathrm{N}.$
• the magnitude of force vector $\stackrel{鈬赌}{\mathrm{C}}\mathrm{is},40\mathrm{N}.$
• The angle made by $\stackrel{鈬赌}{A}$ with x-axis is, $30掳$.
• The angle made by $\stackrel{鈬赌}{B}$ with y-axis is, $30掳$.
• The angle made by $\stackrel{鈬赌}{C}$with x-axis is, $53掳$.

Vector addition is used to sum up two or more vectors together to find the resultant vector.

Let $\stackrel{鈬赌}{D}$be the fourth force is given by,

$\stackrel{鈬赌}{\mathrm{D}}=-\left(\stackrel{鈬赌}{\mathrm{A}}+\stackrel{鈬赌}{\mathrm{B}}+\stackrel{鈬赌}{\mathrm{C}}\right)$

There are two components of $\stackrel{鈬赌}{D}$are $D_X$and $D_y$.

The component ofthe force for theA vector on x-axis is given by,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{X}}=\mathrm{Acos}30掳 \\ =86.6\mathrm{N} \end{gathered}$$

The component of force fortheA vector on y-axis is given by,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{y}}=\mathrm{Asin}30掳 \\ =50.0\mathrm{N} \end{gathered}$$

The components of force for the $\stackrel{鈬赌}{B}$and $\stackrel{鈬赌}{C}$on x-axis and y-axis are given by,

role="math" localid="1655797244543" $$\begin{gathered} {\mathrm{B}}_{\mathrm{X}}=-\mathrm{Bsin}30掳 \\ =-40\mathrm{N} \\ {\mathrm{B}}_{\mathrm{y}}=\mathrm{Bcos}30掳 \\ =69.28\mathrm{N} \\ {\mathrm{C}}_{\mathrm{X}}=-\mathrm{C}\cos 53掳 \\ =-24\mathrm{N} \\ {\mathrm{C}}_{\mathrm{y}}=-\mathrm{C}\sin 53掳 \\ =-31.90\mathrm{N} \end{gathered}$$

Substitute these values in equation for $\stackrel{鈬赌}{D}$and two components of fourth force vector in x and y-axis are given by,

$$\begin{gathered} {\mathrm{D}}_{\mathrm{X}}=-\left({\mathrm{A}}_{\mathrm{X}}+{\mathrm{B}}_{\mathrm{X}}+{\mathrm{C}}_{\mathrm{X}}\right) \\ =-\left(86.6-40-24.07\right)\mathrm{N} \\ =-22.53\mathrm{N} \\ {\mathrm{D}}_{\mathrm{y}}=-\left({\mathrm{A}}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{y}}+{\mathrm{C}}_{\mathrm{y}}\right) \\ =-\left(50+69.28-31.90\right)\mathrm{N} \\ =-87.3\mathrm{N} \end{gathered}$$

The magnitude of $\stackrel{鈬赌}{D}$is given by,

$D=\sqrt{{\left(D_x\right)}^2+{\left(D_y\right)}^2}$

Here, $D_y$is the component of the fourth force vector in y direction and $D_x$ is the component of the fourth force vector in x direction..

Substitute values in the above,

$\mathrm{D}=\sqrt{{\left(-22.53\mathrm{N}\right)}^2+{\left(-87.3\mathrm{N}\right)}^2}$

The direction of the fourth force vector is given by the angle $\left(鈭�\right)$, it makes with the x-axis

$鈭�={\tan }^{-1}\left(\frac{{\mathrm{D}}_{\mathrm{y}}}{{\mathrm{D}}^{\mathrm{x}}}\right)$

Substitute values in the above,

$$\begin{gathered} 鈭�={\tan }^{-1}\left(\frac{-87.3N}{-22.53N}\right) \\ =75.54掳 \\ 蠒=180掳+鈭� \\ =256掳 \end{gathered}$$

Thus, the magnitude of the fourth force vector is$90.2\mathrm{N}$ . And the value of both x and y components of the fourth vector are negative. So, the the fourth force vector acts at an angle of $256掳$, counterclockwise to the positive x-axis.