# 2.2 FREE BODY DIAGRAM

The equilibrium of the bodies which are placed on the supports can be considered if we remove the supports and replace them by the reactions which they exert on the body. In Fig. 2.1 (a), if we remove the supporting surface and replace it by the reaction RA that the surface exerts on the balls as shown in Fig. 2.1 (c), we shall get free-body diagram.

The point of application of the reaction RA will be the point of contact A, and from the law of equilibrium of two forces, we conclude that the reaction RA must vertical and equal to the weight W.

Hence Fig.2.1 (c), in which the ball is completely isolated from its support and in which all forces acting on the ball are shown by vectors, is known a free-body diagram. Hence to draw the free-body diagram of a body we remove all the supports (like wall, floor, hinge or any other body) and replace them by the reactions which these support exert on the body. Also the body should be completely isolated

Problem 2.1. Draw the free body diagram of ball of weight W supported by a string AB and resting against a smooth vertical wall at C as shown in Fig. 2.2(a).

Sol. Given:

Weight of ball = W

The ball is supported by a string AB and is resting against a vertical wall at C.

To draw the free-body diagram of the ball, isolate the ball completely (i.e., isolate the ball from the support and string). Then besides the weight W acting at B, we have two reactive forces to apply one replacing the string AB and another replacing the vertical wall AC. Since the string is attached to the ball at B and since a string can pull only along its length, we have the reactive force F applied at B and parallel to BA. The magnitude of F is unknown.

The reaction RC will be acting at the point of contact of the ball with vertical wall i.e., at point C. As the surface of the wall is perfectly smooth*, the reaction Rc will be normal to the vertical wall (i.e., reaction R, will be horizontal in this case) and will pass through the point B. The magnitude of RC is also unknown. The complete free-body diagram is shown in Fig. 2.2(b).

Problem 2.2. A circular roller of weight 100 N and radius 10 cm hangs by a tie rod AB = 20 cm and rests against a smooth vertical wall at C as shown in Fig. 2.3 (a). Determine: (i) the force F in the tie rod, and (ii) the reaction Rc at point C.

Sol. Given:

Weight of roller,   W = 100 N

Radius of roller,   BC = 10 cm

Length of tie rod,   AB = 20 cm

From MBC we get sin⁡θ=BC/AB=10/20=0.5
θ=sin^(-1) 0.5=〖30〗^o

The free-body diagram of the roller is shown in Fig. 2.3 (b) in which

Rc = Reaction at C

F = Force in the tie rod AB

*The reaction at a perfectly smooth surface is always normal to the surface.

Free-body diagram shows the equilibrium of the roller. Hence the resultant force in x-direction and y-direction should be zero.

For Fx = 0, we get Rc F sin  = 0

or             Rc =F sin                                                  … (i)

Fy = 0, we get 100 – F cos  = 0

or            100 = F cos

=115.47 N. Ans.

Substituting the value of F in equation (i),

Rc = 115.47 × sin 30° = 57.73 N. Ans .

Problem 2.3. Draw the free-body diagram of a ball of weight W, supported by a string AB and resting against a smooth vertical wall at C and also resting against a smooth horizontal floor at D as shown in Fig. 2.4(a).

Sol. Given :

To draw the free-body diagram of the ball, the ball should be isolated completely from the vertical support, horizontal support and string AB. Then the forces acting on the isolated ball as shown in Fig. 2.4(b), will be:

I.            Reaction RC at point C, normal to AC.

II.            Force F in the direction of string.

III.          Weight W of the ball.

IV.          Reaction RD at point D, normal to horizontal surface.

The reactions Rc and RD will pass through the centre of the ball i. e., through point B.

Problem 2.4. A ball of weight 120 N rests in a right-angled groove, as shown in Fig. 2.5(a). The sides of the groove are inclined to an angle of 30° and 60° to the horizontal. If all the surfaces are smooth, then determine the reactions RA and RC at the points of control.

Sol. Given:

Weight of ball,   W = 120 N

Angle of groove   = 900

Angle made by side FD with horizontal = 30°
Angle made by side ED with horizontal == 60°

Angle            FDH = 30° and angle EDG = 60°

Consider the equilibrium of the ball. For this draw the free body diagram of the ball as shown in Fig. 2.5 (b).

The forces acting on the isolated ball will be:

(i)                  Weight of the ball =: 120 N and acting vertically downwards.

(ii)                Reaction RC acting at C and normal to FD.

(iii)              Reaction RA acting at A and normal to DE.

The reactions RA and RC will pass through B, i.e., centre of the ball. The angles made by RA and RC at point B will be obtained as shown in Fig. 2.5(c).

Fig. 2.5

In ∆HDC, ∠CDH = 30° and ∠DCH = 90°. Hence ∠DHC will be 60°. Now in ∆HBL,∠BLH=90°and angle LHB = 60°. Hence ∠HBL will be 30°.
Similarly, ∠GBL may be calculated. This will be equal to 60°.
For the equilibrium of the ball,
ΣFx = 0 and ΣFy = 0
For ΣF_x=0,we have R_C sin⁡〖〖30〗^o 〗-R_A sin⁡〖〖60〗^o 〗=0
or R_C sin⁡〖〖30〗^o 〗=R_A sin⁡〖〖60〗^o 〗
or R_C=R_A×0.866/sin⁡〖〖30〗^o 〗 =1.732 R_A
For ΣF_y=0,we have 120-R_A cos⁡〖〖60〗^o 〗-R_C cos⁡〖〖30〗^o 〗=0
or 120=R_A cos⁡〖〖60〗^o 〗+R_C cos⁡〖〖30〗^o 〗
or 120=R_A cos⁡〖〖60〗^o 〗+R_C cos⁡〖〖30〗^o 〗
=R_A×0.5+(1.732 R_A )×0.866 (∵R_C=1.732 R_A)
= 0.5 RA + 1.5 RA = 2RA
∴ R_A=120/2=60 N. Ans.
Substituting this value in equation (i), we get
Rc = 1.732 × 60 = 103.92 N. Ans.

Problem 2.5. A circular roller of radius 5 cm and of weight 100 N rests on a smooth horizontal surface and is held in position by an inclined bar AB of length 10 cm as shown in Fig. 2.6. A horizontal force of 200 N is acting at B. Find the tension (or Force) in the bar AB and the vertical reaction at C.

Sol. Given :

Weight               W = 100 N

Radius i.e.,   BC = 5 cm

Length of bar,   AB = 10 cm

Horizontal force at   B = 200 N

Let F = Tension in the string AB.
Consider the equilibrium of the roller. For this draw the free body diagram of the roller as shown in Fig. 2.6(b).
The reaction RC at point C will pass through point B.
The tension (or force F) will be acting along the length of the string.
As the roller is in equilibrium in Fig.2.6(b), the resultant force in x-direction and y-direction should be zero.
For Fx = 0, we have F cos θ – 200 = 0

= 230.94 N. Ans.

For   Fy = 0, we have RCW F sin θ = 0

RC = W + F sin θ = 100 + 230.94 × sin 30

= 215.47 N. Ans.

Problem 2.6.Two identical rollers P and Q, each of weight W, are supported by an inclined plane and a vertical wall as shown in Fig. 2.7 (a). Assume all the surfaces to be smooth. Draw the free body diagrams of:

(a) roller Q, (b) roller P and (c) rollers P and Q taken together.

Sol. Given:

Weight of each roller = W
Radius of each roller = R

Identical rollers means the radius of each roller is same.

Hence the line EF in Fig. 2.7(a) will be parallel to surface AB.

Each surface is smooth, hence reaction at the point of contact will be normal to the surface.

Let RA = Reaction at point A

RB = Reaction at point B

RC = Reaction at point C

The two rollers are also in contact at point D. Hence there will be a reaction RD, at the point D.

(a) Free-body diagram of roller Q. To draw the free-body diagram of roller Q, isolate the roller Q completely and find the forces acting on the roller Q. The roller Q has points of contact at B, C and D. The forces acting on the roller Q will be [Refer to Fig. 2.7(b)]:

(i) Weight of roller W.

(ii) Reaction RB at point B.

This will be normal to the surface BA at point B.

(iii) Reaction RC at point C. This will be normal to the vertical surface at point C

(iv) Reaction RD at point D. This will be normal to the tangent at point D.

Fig. 27.7(b)

The reactions RB, RC and RD will pass through the centre E of the roller Q. These three reactions are unknown

(b) Free-body diagram of roller P. Free-body diagram of roller P is shown in Fig. 2.7(c). The roller P has points of contact at A and D. The forces acting on the roller P are:

(i) Weight W

(ii) Reaction RA at point A
(iii) Reaction RD at point D.

The reactions RA and RD will pass through point F, i.e., centre of roller P. These two reactions are unknown. If W is given, then these reactions can be calculated.

(c) Free-body diagram of rollers P and Q taken together. When the rollers P and Q are taken together, then points of contacts are A, B and C. The free-body diagram of this case is shown in Fig. 2.7(d). The forces acting are:

i.            Weight W on each roller

ii.            Reaction RA at point A

iii.            Reaction RB at point B

iv.             Reaction RC at point C.

In this case there will be no reaction at point D.

Problem 2.7. Two identical rollers, each of weight W = 1000 N, are supported by an inclined plane and a vertical wall as shown in Fig. 2.8(a). Find the reactions at the point of supports A,B and C. Assume all the surfaces to be smooth.

Fig. 2.8

Sol. Given:

Weight of each roller = 1000 N

Radius of each roller is same. Hence line EF will be parallel to AB.

Equilibrium of roller P. First draw the free-body diagram of roller P as shown in Fig. 2.8 (c). The roller P has points of contact atA and D. Hence the forces acting on the roller P are:

(i) Weight 1000 N acting vertically downward.
(ii) Reaction RA at point A. This is normal to OA.

(iii) Reaction RD at point D. This is parallel to line OA.

The resultant force in x and y directions on roller P should be zero.

For Fx = 0, we have

RD sin 60° -RA sin 30° = 0 or RD sin 60° = RA sin 30°

RD cos 60° + RA cos 30° – 1000 = 0
(0.577 RA) cos 60° + RA cos30° = 1000                       (RD = 0.577 RA)

or 0.577 × 0.5 R A + RA × 0.866 = 1000

1.1545 RA = 1000 or RA = 1.1545 = 866.17 N. Ans.

Substituting this value in equation (i), we get

RD = 0.577 × 866.17 = 499.78

Equilibrium of roller Q. The free-body diagram of roller Q is shown in Fig. 2.8 (b).

The roller Q has points of contact at B, C and D.

The forces acting on the roller Q are :
(i) Weight W = 1000 N ;

(ii) Reaction RB at point B and normal to BO ;

(iii) Reaction RC at point C and normal to CO ; and
(iu) Reaction RD at point D and parallel to BO.

For Fx = 0, we have

RB sin 30° + RD sin 60° – RC = 0

or    RB × 0.5 + 449.78 × 0.866 – RC= 0

or RC= 0.5 RB + 432.8                                                        (ii)

For Fy = 0, we have

RB × cos 30° – 1000 – RD × cos 60° = 0

or RB × 0.866 – 1000 – 499.78 × 0.5 = 0

Substituting this value in equation (ii), we get

RC = 0.5 × 1443.3 + 432.8 = 1154.45 N. Ans.

Problem 2.8.Two spheres, each of weight 1000 Nand of radius 25 cm rest in a horizontal channel of width 90 cm as shown in Fig. 2.9. Find the reactions on the points of contact A, B and C.

Sol. Given:

Weight of each sphere,  W = 1000N
Radius of each sphere,  R = 25 cm

AF = BF = FD = DE = CE = 25 cm

Width of horizontal channel = 90 cm

Join the centre E to centre F as shown in Fig. 2.9(b).

Now EF = 25 + 25 = 50 cm, FG = 40 cm

Equilibrium of Sphere No.2. The sphere 2 has points of contact at C and D

Let RC = Reaction at C

and RD = Reaction at D

The free body diagram sphere No.2 as shown in Fig 2.9(c)

The reaction RD at point D, will pass through the centre E of the sphere No.2, as any line normal to any point of the circumference of the circle will pass through the centre of circle. For the equilibrium of the sphere No.2, the resultant force in x and y direction should be zero.

For = 0, we have RD sin θ = RC
For
= 0, we have RD cos θ = 1000

Equilibrium of sphere No.1.The sphere 1 has points of contact at A, B and D.

Let RA = Reaction at point A

RB = Reaction at point B

The free-body diagram of sphere No.1 is shown in Fig. 2.9(d).

The reactions RA RB and RD will pass through the centre F of the sphere No. l.