Question

Three ants P, Q and R are pulling a grain with forces of magnitude 6N, 3√3N and 3√2N as shown in the
figure. Find the magnitude of resultant force (in N) acting on the grain.​

Answer 1

Explanation:

the net force acting at the centre is 3N along positive x axis

Answer 2

Answer:

The magnitude of the resultant force is 3.718 N

It is acting at a angle 36.2° with positive X axix

Explanation:

6N force is making 30° angle with x axis.

3√2 N force is making -45° angle with x axis.

And 3√3 N force is towards negetive (-ve) x axis.

So if we divided each forces into X and Y axis components; We will get:

X axis components:

$x = 6 \cos30 \degree + 3 \sqrt{2} \cos( - 45 \degree) + 3 \sqrt{3} \cos180 \degree \\ x = (5.196 + 3 - 5.196) N\\ x = 3 \: N$

Y axis components:

$Y = 6 \sin60 \degree + 3 \sqrt{2} \sin( - 45 \degree) \\ Y = 2.196 \: N$

So the Resultant Force (Fr) = Fx + Fy :

$|Fr| = \sqrt{ {(Fx)}^{2} + {(Fy)}^{2} } \\ |Fr| = \sqrt{ {3}^{2} + {2.196}^{2} } \: N \\ |Fr| = 3.718 \: N$

So the magnitude of the resultant force is 3.718 N.

Direction of this resultant force;

$\theta = \ { {tan}^{ - 1} ( \frac{Fy}{Fx })} \\ \theta = { {tan}^{ - 1} ( \frac{2.196}{3})} \\ \theta = 36.2 \degree \: with \: x \: axis$

So this resultant force is acting at a angle 36.2° with positive X axix