We have learnt that a force can be replaced by a force and a couple. When a number of forces are acting on a body in space then each force is transferred to an arbitrary reference point along with a couple for each force transferred. Thus, we get

*(i) *a system of concurrent forces and

*(ii) *a system of couples.

.The system of concurrent forces can be added to get the resultant single force **(R)**. Also the system of couples can be added to get the resultant moment **(M)**.

Hence the equations become as

**R=F**_{1}**+F**_{2}**+F**_{3}**+……**

**M=M _{1}+M_{2}+M_{3}**

**+**

**……**

An equivalent system for a given system of coplanar forces, is a combination of a force passing through a given point and a moment about that point. The force is the resultant of all forces .

**Problem 1.53**. *Three external forces are acting on a **L-shaped **body as shown in Fig.1.87. Determine the equivalent system through point *O.

**Sol**. Given:

Force at *A *= 2000 N, Angle = 30°

Force at *B *= 1500 N

Force at C = 1000 N

Distance *OA *= 200 mm, *OB *= 100 mm and *BC=200mm *

Angle *COA *= 90°

acting on the body. And the moment is the sum of all the moments about that point.

Hence equivalent system consists of:

*(i) *a single force *R *passing through the given point *P *and

*(ii) *a single *moment M*_{R}* *

where *R *= the resultant of all force acting on the body.

*M*_{R}* *= sum of all moments of all the forces about point *P. *

**Fig. 1.87**

Determine the equivalent system through O. This means find

*(i) s*ingle resultant force, *R *

*(ii)* single moment through O.

Taking *x*-axis along *OA *and *y*-axis along OC

The force at A is resolved into two components.

Component along x-axis = 2000 × cos 30o = 1732 N

Component along y-axis = 2000 × sin 30o = 1000 N

Resolving all forces along x-axis i.e.,

ƩFx = 2000 cos 30° – 1500 – 1000 = – 768 N

Similarly ƩFy = – 2000 × sin 30° = – 1000 (- ve sign is due to downward)

∴Resultant, R=√(ƩF_(x^2 )+ƩF_(y^2 ) )=√(〖(-768)〗^2+〖(-1000)〗^2 )

= √(589824+1000000)=1260.88 N

Taking moments of all forces about point O,

M0 = (- 2000 sin 30) × 200 + 1500 × 100 + 1000 × 300

= – 200000 + 150000 + 300000

= 250000 Nmm = 250 Nm

∴ Equivalent system through point O is

R = 1260.88 N

M=250Nm.

**Problem 1.54**. *Fig. *1.88 *shows two vertical forces and a couple of moment 2000 Nm acting** **on **a horizontal rod which is fixed at end A.*

*(i) **Determine the resultant of the system. *

*(ii) **Determine an equivalent system through A. *

**Sol**. Given:

Force at C = 4000 N, Force at *B *= 2500 N

Moment at *D *= 2000 Nm

Distance *AC *= 1 m, *BC *= 1.5 m

*CD=0.8m, BD=0.7m
*

*(i) Re*

*sultant of the system*

This means to find the resultant of all the forces and also the point at which the resultant is acting. There are two vertical forces only.

Hence resultant, *R *= 4000 – 2500 = 1500 N acting downward

The point at which the resultant is acting is obtained by taking moments about point A. For moments there are two forces (4000 N at C and 2000 N at *B) *and also a moment at D.

**Fig. 1.88**

Moment of force 4000 N about point *A *= 4000 × 1

= 4000 Nm (clockwise)

Moment of force 2500 N about point *A *

= 2500 × (1 + 1.5) = 2500 × 2.5

= 6250 Nm (anti-clockwise)

Moment at *D *= 2000 Nm (clockwise)

Sum of all moments about *A *

= 4000 (clockwise) – 6250 (anti-clockwise) + 2000 (clockwise)

= – 250 (anti-clockwise)

The resultant is acting vertically downward. If it is acting towards right of *A, *then it will give cloakwise moment. But we want anti-clockwise moment. Hence the resultant must act towards the left of A.

Let *x *= Distance of resultant force (1500 N) from *A
* Moment of resultant force

*(R)*about A

= 1500 *× x *

1500 × *x *= 250

Hence resultant of the system is 1500 N↓ acting at a distance *of 0.166 *m left *to A. ***Ans. **

*(ii) **Equivalent system through A *

This means to find a single resultant force and a single moment through *A. *

Single resultant force, *R *= 1500 N

Single moment through, *A *= 250 Nm. Ans.

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