Question

2 forces whose magnitude are in the ratio 3:5 gives a resultant of 35N if the angle of inclination is 60 degree calculate the magnitude of force

Answer 1

Answer:

The magnitude of forces will be 15N and 25 N respectively.

Explanation:

Let the net force be "F". And the two forces be F₁ and F₂ respectively.

As per the question we have,

$\frac{F_1}{F_2}=\frac{3}{5}$

$F_1=\frac{3F_2}{5}$             (1)

So,

$F=\sqrt{F^2_1+F^2_2+2F_1F_2cos\theta}$            (2)

The values that are  given in the question,

F=35N

θ=60°

By substituting the required values in equation (2) we get;

$35=\sqrt{(\frac{3F_2}{5})^2+F^2_2+2\times \frac{3F_2}{5}\times F_2\times cos60\textdegree}$            ($cos60\textdegree=\frac{1}{2}$)

$(35)^{2}=\frac{9F^2_2}{25}+F^2_2+\frac{3F^2_2}{5}$

$(35)^{2} =\frac{49F^2_2}{25}$

$F^2_2=\frac{(35)^2\times 25}{49}$

$F_2=\sqrt{\frac{(35)^2\times 25}{49}} =25N$                    (3)

By putting equation (3) in equation (1) we get;

$F_1=\frac{3\times 25}{5}=15N$

Hence, the magnitude of forces acting at the inclination of 60° will be 15N and 25 N respectively.