91Ó°ÊÓ

The given data is listed below,

• The angle at which a larger pull is directed is, $\mathrm{θ}=21°$west to north
• The resultant pull directed northward is, $N_P=460\mathrm{N}$.

The scalar coefficients are termed components when a vector is uniquely represented as a linear combination of special vectors derived from a basis.

The resultant of two component of force is given by,

$\stackrel{→}{A}+\stackrel{→}{B}=\stackrel{→}{C}$

Here, ${\stackrel{→}{A}}_x$} is the vector component of force A, $\stackrel{→}{B}$is the vector component of B and$\stackrel{→}{C}$is the resultant of two vector components.

The magnitude of the smaller pull and its direction for two unknown quantities is given by the two equations given below,

$$\begin{gathered} {\stackrel{→}{A}}_x+{\stackrel{→}{B}}_x={\stackrel{→}{C}}_x \\ {\stackrel{→}{A}}_y+{\stackrel{→}{B}}_y={\stackrel{→}{C}}_y \end{gathered}$$

Here, ${\stackrel{→}{A}}_x$is the horizontal vector force A, ${\stackrel{→}{B}}_x$is the horizontal vector of force B and ${\stackrel{→}{C}}_x$is the resultant of the two components of force A and B. and

The value of $B_x$is width="110">Bx=-Bsin21°and $B_y$ is given by $B_y=B\sin 21°$

Let smaller pull be $\stackrel{→}{A}$ and larger will be $\stackrel{→}{B}$. Hence,

$B=2A$

Substitute value in resultant formula given by,

$$\begin{gathered} A\cos \mathrm{θ}\hat{i}+\left(-B\sin 21°\right)\hat{i}=0 \\ 2A\sin 21°=A\cos \mathrm{θ} \\ \mathrm{θ}=44.21° \end{gathered}$$

So, the value of$\mathrm{θ}$is $44.21°$.

The equation for y-direction gives values as,

$$\begin{gathered} 2A\cos 21°+A\sin \mathrm{θ}=460\mathrm{N} \\ A\sin 44.21°+2A\cos 21°=460\mathrm{N} \\ 2.5645A=460\mathrm{N} \end{gathered}$$

So, the value of A is given by,

$A=179.37\mathrm{N}$

$\stackrel{→}{A}$must have an eastward component to cancel the westward component of$\stackrel{→}{B}$These are concluded from the figure given below,