# Solved Examples- Concept of Equilibrium of Rigid Body

SOLVED EXAMPLES

Example 4.1. A circular roller of weight 1000 N and radius 20 cm hangs by a tie rod AB = 40 cm and rests against a smooth vertical wall at C as shown in Fig. P-4.1(a). Determine (i) the force F in the tie rod ; (ii) the reaction at point C.

Solution. Let us draw the free-body diagram of the circular roller.

Sin θ =20/40 =1/2  ; θ = 300

Applying conditions of equilibrium

Ʃ f x = 0

-f sin θ + R c =0

Ʃ f y = 0 => f cos θ -1000  = 0

From (2)          f = 1000 /cos 30= 1154 .7 N

From (1)          RC =F sin 300  = 1154.7 sin 300 =577.35 N

Example 4.2: A ball of weight 1200 N rests in a right angled groove as shown in Fig. P-4.2(a). surfaces are smooth. Find out the reactions at A and C.

Solution.

Let us draw the free body diagram. It is shown in Fig. 4.2(b)

Ʃ Fx  = 0

Ʃ fx = 0 ; RC sin  600  +RA sin 300 -1200 = 0

From (2) =>                   RC x √3 / 2 + RA /2  =1200

Or                                √ 3 RC +RA  =2400

From (1)                      RC / 2 – RA √3/2 = 0

Or                                 RC √3 RA

From (3) =>                 √3 x √3 RA +RA  =2400

RA =2400/4 =600 N.

RC =√3 RA =√3 x 600 = 1039 .2 N .

Example 4.3. A circular roller of radius 10 Cm and of weight 1000 N rests on a smooth horizontal surface and is held in position by an inclined bar AB of length 20 Cm as shown in Fig. 4.3(a). A horizontal force of 2000 N is acting at B. Find the tension in the bar AB and the  Vertical reaction at C.

Solution :

Cos θ =10 /20 =1/2

θ =600

Ʃ f x = 0 => = T sin θ +2000 = 0

T = 2000 / sin θ = 2000 / sin 600 =2309.4 N

Ʃ FY = 0

Or                    -T cos θ + RC -1000 = 0

Or                    R C =1000 +T cos 600

= 1000 + 2309.4 cos 600

=2154.7 N

Example 4.4. Two identical rollers, each of weight 100 N, are supported by an inclined plane and a vertical plane as shown in Fig. P-4.4(a). Find the reactions at the points of supports A, B and C. All surfaces are smooth

Solution. Free body diagram of rollers are as follows : Refer (Figure P-4.4(b) & (c))

For roller E (fig . p- 4.4 ©

Ʃ f x = 0 RC –RB cos 600 –RD  sin 600 = 0

Ʃ f y = 0  RB sin 600 -100 –RD cos 600 = 0

For roller f (fig -4.4 ©

Ʃ f x = 0   => RD cos – RA cos 600 = 0

Ʃ f y = 0  => RD sin 300 + RA sin 600 -100 = 0

From (3)                      RD =RA cos 600 / cos 300

From (4) =>                 RA cos 600 sin 300 / cos 300 +RA sin 600 =100

RA = 86.6 N

RD = 86.6 x cos 600 /cos 300 = 50 N

From (2) =>                 RB sin 600 = 100 + RD cos 600

=125

Or                                 RB =125 / sin 600 =144.34 N

From (1) =>                 RC  =RB cos 600 +RD sin 600

=144.34 cos 600 +50 sin 600

= 115.47 N

Example 4.5. Two spheres each of weight 500 Nand of radius 50 Cm rest in a  Horizontal channel of width 180 em as shown in Fig. P-4.5(a). Find the reactions on the points of contact A, Band C.

Solution .

Sin θ = 80 /100 =4/5

And                             cos θ = 3/5

For sphere f

Ʃ f x =0 => -RA –RD sin θ

Ʃ f y = 0 =>  -RD cos θ =0

For sphere E

Ʃ f x  = 0 => -RC +RD sin θ = 0

Ʃ f y = 0 => RD cos θ -500 = 0

From (4) =>                 RD =500 /cos θ =500 x 5 /3 833. 34 N

From (3) =>                 RC =RD sin θ = 833.34 x 4/5

=666.72 N

From (2) =>                 RB =RD cos θ +500

=1000 N

From (1) =>                 R= =RD sin θ = 833 .34 x 4/5 =666.672 N

Example 4.6. Two smooth circular cylinders, each of weight 100 N and radius 30 cm, are connected at their centres by a string AB of length 80 cm and rest upon a horizontal plane, supporting above them a third cylinder of weight 200 N and radius 30 cm as shown in Fig. P- 4.6(a). Find the tension Tin the string AB and the pressure produced on the floor at the points of contact D and E.

Solution.      Sin θ =40 /60 -2 /3

Cos θ =

=√ 5 /3

Fig. P-4.6(b) (Free body diagrams of cylinders)

For sphere A,

Ʃ  F x =0 ; T – R F sin θ =0

Ʃ f y =0 ;  – RF cos θ -100 = RD =0

For sphere B,

Ʃ f x  =0 , RG sin θT = 0

Ʃ F Y = 0 ,T RG cos θ + RE -100 =0

For sphere C,

Ʃ  f x  =0 ,RF sin θRG sin θ = 0

Ʃ F Y = 0 ,RF cos θ + RG cos θ -200 =0

From  (5) =>                            RF =RG

From (6) =>                 2 RF cos θ =200

Or                                RF =100 /cos θ =100  x 3 / √ 5 =1341.16 =RG

From (1) =>                 T= RF sin θ =1341.16 x 2/3 = 89.44

From (2) =>                  RD = RF cos θ  +100 =200 N

From (1) =>                 RE =100 +R=  cos θ =200 N .

Example 4.7. A roller of radius 20 cm weighing 300 N is to be pulled over a   rectangular block of height 10 cm as shown in Fig. P-4.7(a) by a horizontal force applied at the end of a string wound the circumference of the roller. Find the magnitude of the horizontal force which will just tum the roller over the corner of the rectangular block. Also determine the magnitude and direction of reaction at A and B. All surfaces are smooth.

Solution.                 Cos 2 θ =1/2 =cos 600

θ =300 , sin 2θ =DB/20  or  DB  =20  sin 2θ =20 sin 600

Taking moment about B, we get,

– F x 30 + W x 20 sin 60°- RA x 20 sin 60° = 0

RA = 0, because roller will detach from A.

For just turning about B .

F x 30 = 20 W sin 60°

F = 20 x 300 sin 60° /30

= 200 x √3/2 = 173.205 N

For reaction at B

Ʃ Fy  =0 => RB cos θ  -W =0

RB = W /cos θ  =300 / cos 300  =346.41 .

RA =0 .

Example 4.8 . With what force magnitude F must the person pull on the cable in order to cause the scale A to read 2000  N ? The weights of the pulleys and cables are negligible. State any assumptions :

Solution .                        500  x  9.81 -5F= 2000 N

5F= 2905

F =2905 /5 581 N

Assumptions:

1. Pulleys are frictionless

2. All the tensions are vertical.

3. Strings are inextensible.

Example 4.9. The 500 kg uniform beam is subjected to the three external loads as shown in Fig. P-4.9 (a). Calculate the reactions  at the support point 0. The x-y plane is vertical.

Solution:

Ʃ F y  =  0 ,ROX -3 cos 600 =0

ROX =3 cos 600 = 1.5 KN .

Ʃ F Y =0 => ROY +1.4 -500  x 9.81/1000 -3 cos 300 =0

ROY =6.103 KN.

Ʃ MO =0 => -M +1.4 x 1.2 -500 x 9.81 /1000 x 2.4 +15 – cos 300 x 4.8 =0

-M =7.5627 .KNm .

M = -7.5627 KNm .

=7562.7 NmCCW

Example 4.10. A simply supported beam AB of length 4.5 m, carries a uniformly distributed load of 20 kN/m for a distance of 3 m from the left end. Determine the reactions at A and B.

Solution:

Ʃ  F Y =0

RA -60 +RB =0

RA + RB =60

Ʃ MA =0

-60 x 1.5 +RB x  4.5 =0

Or        RB =60 x 1.5/4.5 =20 KN.

(1) =>   RA =60 –RB =60 -20 -40 KN.

Example 4.11. A simply supported beam of span 18m carries a uniformly varying load from zero at end A to 450 N/m at end B. Find the reactions at the two ends of the support.

Solution:      Area of ∆ ABC =1/2 x 18 x 450

=4050 N

This load will act at CG of the triangle ABC i.e., at a distance of 2/3 x AB = 2/3 x 18 =3  12 m from end A.

Ʃ F y =0

RA +RB -4050 =0

RA =RB =4050

Ʃ MB = 0

-RA x 18 +4050 x 6 =0

Or            RA = 4050  x  6/ 18 =1350 N.

(1) =>           RB =4050 -1350 =2700 N.

Example 4.12. A simply supported beam of length 10 m carries a uniformly increasing load of 400 N/m at one end to 800 N/m at other end. Determine the  reactions at both ends.

Solution. Area of rectangle ABEC = 400 x 10 = 4000 N acting at a distance 5 m from A.

Area of the triangle    EDC = 1/2 x 10 x 400

= 2000 N

acting at a distance 2/3  x 20 /3 m =6.67 m from A.

RA +RB =6000 N

Ʃ MA =0 => RB x  10 -4000 x 5 -2000 x 6.67 =0

RB = 3334 N.

RA =6000 -3334 =2666N.

Example 4.13. Use method of joints to find forces in the members AB, AC and BC of the truss shown in Fig. P-4.13.

Solution.  First we find out RB and RC.

For ∆ABC, cos 60° = AB /5 ; AB = 5 x 1/2 =2.5 m

For  ∆ABC ,cos 600  =BD/AB , ; BD =1/2 x 1/2 x 2.5 =1.25m.

Taking moment about C, we get

– 5RB + 200 (5 – 1.25) = 0

RB =200)5-1.25) /5 = 40 (5 -1.25) N

= 150N.

and                             RB +RC= 200 N

or                                                . RC= 200-150 = 50 N.

Then we consider the equilibrium of various joints.

For joint B

Ʃ F x = 0

FBC –FAB cos 600 =0

2 FBC =FAB x 1/2

Ʃ FY =0 => RB –FAB sin 600 =0

FAB =RB /sin 600 = 150 /sin 600 =173.2 N.(Compressive)

(1)                                              FBC =1/2 FAB =1/2 x 173.2 =86.6 N. (tensile)

For joint C:

Ʃ F x  = 0

-FCB +FAB cos 300 = 0

Or                                    FCB =FAB cos 300

We have,                        FCB =FAB

FAC =FCB /cos 300 =FBC /cos 300 = 86.6 /cos 300 =100 N  (compressive).

Example 4.14. Use method of sections to find the forces in the numbers AB and AC of the truss shown in Fig. P-4.14.

Solution. In this method, a section line is passed through the members, in which forces are to be determined. The section line should be drawn in such a way that it does not cut more than three members of unknown forces acting on it.

Let us draw a section (1)- (1) cutting the members AB and BC in which forces are to be determined. Consider the equilibrium of left part of the truss.

Taking moment of all the forces acting on the left part about C, we get

-150 x 5 –FBA x AC = 0

FBA = – 150 x 5 /ac = – 150 x 5 /5 cos 300 = -150 /cos 300  =- 173.2 N.

The negative sign shows that it is compressive (opposite to the direction taken)

FAB= 173.2 N.

Again taking moment about A, we get

-150 x 1.25 +FBC x AD =0

FBC = 150 x 1.25 /AD = 150 x  1.25 /2.5 sin 600 = 86.6 N. (tensile)