Question

two forces 8N and 12N act at 120 degrees.the third force required to keep this in equilibrium

Answer 1

Answer:

$\large\boxed{\sf{-4\sqrt{7}\;N}}$

Explanation:

Given that, two forces are there.

• f1 = 8 N
• f2 = 12 N

Angle between them = 120°

Now, to find the third force for keeping these two in equilibrium.

We need to find the resultant of these two forces.

Therefore, we will get, net force,

$= > f_{net} = \sqrt{ {f_{1}}^{2} + {f_{2} }^{2} + 2f_{1}f_{2} \cos120 }$

Substituting the values, we get,

$= > f_{net} = \sqrt{ {(8)}^{2} + {(12)}^{2} + 2 \times 8 \times 12 \times \frac{ - 1}{2} } \\ \\ = > f_{net} = \sqrt{64 + 144 - 96} \\ \\ = > f_{net} = \sqrt{112} \\ \\ = > f_{net} = \sqrt{16 \times 7} \\ \\ = > f_{net} = 4 \sqrt{7}$

Therefore, net force is 4√7 N.

Now, the third required force will be just in opposite direction of net force having same magnitude.

Hence, required force = -4✓7 N

Answer 2

Answer:

we have to find the resultant force

Explanation:

(Fres)^2=(F1)^2+(F2)^2+2*F1*F2*COSX(HERE X IS THE ANGLE)

(Fres)^2=64+144+2*8*12*COS120

(Fres)^2=208+16*12*(-0.5)

(Fres)^2=208-96

(Fres)^2=112

Fres=√112

Fres=10.58N