Question

The resultant of two forces acting at an angle of ${150}^{\circ }$ is $10kgwt$, and is perpendicular to one of the forces. The smaller force is :

• A. $10\sqrt{3}kgwt$
• B. $20\sqrt{3}kgwt$
• C. $20kgwt$
• D. $10kgwt$

Answer 1

The correct option is A $10\sqrt{3}kgwt$

From, the diagram you can tell the other force (magnitude $B$) must be act at ${150}^o$ clockwise from $A$ (i.e. ${60}^o$ clock wide from the $x-$axis).

You can draw a vector adding triangle from $A+B.R$, but using components.

In the $X-$ direction:

$B.\cos \left({60}^o\right)=10$

Since $\cos \left({60}^o\right)=1.2$

$B\left(1.2\right)=10$

$B=\frac{10}{1.2}$

$\therefore B20kgwt$

In the $y-$direction:

$A=B\left(\sin {60}^o\right)$

Since $\sin {60}^o\sqrt{3/2}$

$A=20\times \sqrt{3}/2$

$A=10.\sqrt{3}kgswt$

$\therefore 10\sqrt{3}$ is smaller than $20$.

So, the answer is $10\sqrt{3}kgswt$.