Q60P - University Physics with Modern Physics

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.

                                         

Step-by-Step Solution

Answer

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

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 of the force for the A 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 for the A 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,

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

 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,

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

 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.