1.6 Components of a vector  (Page 4/4)

$$F_x=10\sin {30}^0=10X\frac{1}{2}=5N$$ $$F_y=10\cos {30}^0=10X\frac{\sqrt{3}}{2}=5\sqrt{3}N$$ $$\mathbf{F}=-F_x\mathbf{i}-F_y\mathbf{j}=-5\mathbf{i}-5\sqrt{3}\mathbf{j}N$$

$$F_x=10\sin {30}^0=10X\frac{1}{2}=5N$$ $$F_y=10\cos {30}^0=10X\frac{\sqrt{3}}{2}=5\sqrt{3}N$$ $$\mathbf{F}=F_x\mathbf{i}-F_y\mathbf{j}=5\mathbf{i}-5\sqrt{3}\mathbf{j}N$$

From graphical representation, the angle that vector makes with y-axis has following trigonometric function :

Angle

$$\begin{array}{c}\tan \theta =\frac{a_x}{a_y}\end{array}$$

Now, we apply the formulae to find the angle, say θ, with y-axis,

$$\begin{array}{c}\tan \theta =-\frac{\surd 3}{1}=-\surd 3=\tan 120°\end{array}$$

$$\begin{array}{c}\Rightarrow \theta =120°\end{array}$$

In case, we are only interested to know the magnitude of angle between vector and y-axis, then we can neglect the negative sign,

$$\begin{array}{c}\tan \mathrm{\theta ’}=\surd 3=\tan 60°\end{array}$$

$$\begin{array}{c}\Rightarrow \mathrm{\theta ’}=60°\end{array}$$

The vector can be expressed in terms of its component as :

$$\begin{array}{c}\mathbf{a}=a_x\mathbf{i}+a_y\mathbf{j}+a_z\mathbf{k}\end{array}$$

where $a_x=a\cos \alpha ;a_y=a\cos \beta ;a_z=a\cos \gamma$ .

Angles

The magnitude of the vector is given by :

$$\begin{array}{l}a=\surd ({a_x}^2+{a_y}^2+{a_z}^2)\end{array}$$

Putting expressions of components in the equation,

$${\displaystyle \begin{array}{l}\Rightarrow a=\surd (a^2{\cos }^2\alpha +a^2{\cos }^2\beta +a^2{\cos }^2\gamma )\end{array}}$$

$${\displaystyle \begin{array}{l}\Rightarrow \surd ({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma )=1\end{array}}$$

Squaring both sides,

$${\displaystyle \begin{array}{l}\Rightarrow {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1\end{array}}$$

The component of the weight along the incline is :

Weight on an incline

$$\begin{array}{c}{W}_x=\mathrm{mg}\sin \theta =100x\sin 30°=100x\frac{1}{2}=50N\end{array}$$

and the component of weight perpendicular to incline is :

$$\begin{array}{c}{W}_y=\mathrm{mg}\cos \theta =100x\cos 30°=100x\frac{\surd 3}{2}=50\surd 3N\end{array}$$

We depict the situation as shown in the figure. The resultant force R is shown normal to small force $F_1$ . In order that the sum of the forces is equal to R, the component of larger force along the x-direction should be equal to smaller force :

Two forces acting on a point

$$\begin{array}{c}{F}_2\sin \theta =F_1\end{array}$$

Also, the component of the larger force along y-direction should be equal to the magnitude of resultant,

$$\begin{array}{c}{F}_2\cos \theta =R=8\end{array}$$

Squaring and adding two equations, we have :

$$\begin{array}{c}{F}_2^2=F_1^2+64\end{array}$$

$$\begin{array}{c}\Rightarrow F_2^2-F_1^2=64\end{array}$$

However, according to the question,

$$\begin{array}{c}\Rightarrow F_1+F_2=16\end{array}$$

Substituting, we have :

$$\begin{array}{c}\Rightarrow F_2-F_1=\frac{64}{16}=4\end{array}$$

Solving,

$$\begin{array}{c}\Rightarrow F_1=6N\\ \Rightarrow F_2=10N\end{array}$$