Question

find the magnitude and direction of the resultant force acting on the body in a given figure​

Answer 1

Answer:

• Magnitude of Resultant ,R = 121.65
• Direction of Resultant is 30√3/100 along the direction of F2.

Explanation:

According to the Question

It is given that,

• $\sf\: F_1 = 60N$
• $\sf\:F_2 = 80N$
• $\sf\: Angle~ between ~ them, \theta = 60^{\circ}$

We have to calculate the magnitude of resultant & its direction .

Let the Resultant be R

$\bf\:R = \sqrt{F_1^2 + F_2^{2} + 2F_1F_2Cos\theta}$

by putting the value we get

$\tt\implies R = \sqrt{60^2 + 80^2 + 2.60.80. Cos60^{\circ}} \\\\\tt\implies R = \sqrt{3600 + 6400 + 9600 . 0.5} \\\\\tt\implies R =\sqrt{3600 + 6400 + 4800} \\\\\tt\implies R = \sqrt{14800} \\\\\tt\implies R =121.65$

• Hence, the magnitude of resultant is 121.65 .

NowNow calculating the direction of resultant,

• $\bf\:tan\alpha = \frac{F_1Sin\theta}{F_1 + F_2 Cos\theta}$

substituting the value we get

$\tt\implies tan\alpha = \sf\frac{60Sin60^{\circ}}{60 + 80 Cos60^{\circ}} \\\\\tt\implies tan\alpha = \frac{30\sqrt{3} }{70}$

Hence, the direction of the resultant is along the direction of F2 .

Answer 2

Answer: Resultant force= 121.65N

Direction= It makes an angle of 25.28° with the direction of 80N force.

Explanation:

Let the resultant force be R and the angle between the resultant force and the 80N force be theta.

Therefore:

R² = 60²+80²+2×80×60×cos60°

R = 121.65N