Question

Three horizontal ropes pull on a large stone stuck in the ground, producing the vector forces $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$, $\stackrel{\rightharpoonup }{\mathbf{B}}$, and$\stackrel{\rightharpoonup }{\mathbf{C}}$ shown in Fig. P1.60. Find the magnitude and direction of a fourth force on the stone that will make the vector sum of the four forces zero.

Short answer

The fourth force of $90.2\mathrm{N}$acts at an angle of $256°$ ,counterclockwise to the positive x-axis.

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of a vector quantity.

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

Determination of the magnitude and direction of a fourth force on the stone that will make the vector sum of the four forces zero

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

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

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

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

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

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

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

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

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

Substitute these values in equation for $\stackrel{\rightharpoonup }{D}$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 $\stackrel{\rightharpoonup }{D}$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 $\left(\propto \right)$, it makes with the x-axis

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

Substitute values in the above,

$$\begin{gathered} \propto ={\tan }^{-1}\left(\frac{-87.3N}{-22.53N}\right) \\ =75.54° \\ \varphi =180°+\propto \\ =256° \end{gathered}$$

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 $256°$, counterclockwise to the positive x-axis.