Question

Two forces whose magnitudes are in the ratio $3:5$ give a resultant of $28\text{N}$. If the angle of their inclination is ${60}^{\circ }$, find the magnitude of the larger force.

• A. $24\text{N}$
• B. $28\text{N}$
• C. $32\text{N}$
• D. $20\text{N}$

Answer 1

The correct option is D $20\text{N}$

Let $A$ and $B$ be the two forces.
Let $A=3x;B=5x;R=28\text{N}$ and $\theta ={60}^{\circ }$
We know that $R=\sqrt{A^2+B^2+2AB\cos \theta }$
$\Rightarrow 28=\sqrt{(3x{)}^2+(5x{)}^2+2\left(3x\right)\left(5x\right)\cos {60}^{\circ }}$
$=\sqrt{9x^2+25x^2+15x^2}=7x$
$\Rightarrow x=\frac{28}{7}=4$
Hence, forces are $A=3\times 4=12\text{N},B=5\times 4=20\text{N}$