When the force system gets its resultant along the line of action of force in the same plane, it is called equilibrant. Similar arbitrary point can also be shown for moment of forces about the point of resultant of force. Thus, it is called the equivalent system of forces that can be formed because of two reasons-

**System of concurrent forces**

Let us consider three forces F_{1}, F_{2} and F_{3}act on a rigid body in different direction such that force F_{1} and F_{2} have different directions than force F_{3}.

Then,

Resultant force can be given by-

R = F_{1} + F_{2}+ F_{3} + …… (In the direction of force F_{3})

**System of couples**

Similarly, if the moment of forces F_{1}, F_{2} and F_{3} are considered such that resultant of moment of force give the direction about point of resultant of force, then it can be given by-

M = M_{1} + M_{2} + M_{3} + ……..

**Problems on Equivalent Systems of Forces**

**Example:**

Fig. 1.87 shows L-shaped body. Find the magnitude of three external forces which are acting on it such that the body will have equivalent system through point O.

**Solution:**

**Given Data:**

Force at A, F1 = 2000 N

Force at B, F2 = 1500 N

Force at C, F3 = 1000 N

OA = 200 mm

OB = 100 mm

BC = 200 mm

Force, F1 making an angle 300with the X-axis as shown in above figure

We consider,

Resultant R will be passing through a point and moment MR

Now,

Components ofF1 can be given by-

Along x-axis, F1x = 2000 x cos 300= 1732 N

Along y-axis, F1y = 2000 x sin 300= 1000 N

According to the equilibrium of forces, we can write-

∑▒〖F_x=〗1732 – 1500 – 1000

= – 768 N

∑▒〖F_y=〗 – 1000N

Resultant can be given by-

R = √(∑▒F_x 2+ ∑▒F_y 2)

= √[(– 768)2+ (– 1000)2]

R = 1260.88 N

Now,

Moment of forces at point O can be given by-

M = – 1000 x 200 + 1500 x 100 + 1000 x 300

= 250000 Nmm

Or,

M = 250 Nm