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1.15. EQUIVALENT SYSTEMS OF FORCES

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=F1+F2+F3+……

M=M1+M2+M3+……

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 MR

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

MR = sum of all moments of all the forces about point P. 

Fig.-1.87

Fig. 1.87

Determine the equivalent system through O. This means find

(i) single 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) Resultant 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

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.