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 R_{A} 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 R_{A} will be the point of contact A, and from the law of equilibrium of two forces, we conclude that the reaction R_{A}* *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 *R _{C} *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

*R*is also unknown. The complete free-body diagram is shown in Fig.

_{C}*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 sinθ=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 **F _{x} *= 0, we get

*Rc*–

*F*sin = 0

*or Rc **=F *sin * … (i)*

*F _{y} *= 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 *R _{C} *at point

*C,*normal to

*AC.*

II. Force *F *in the direction of string.

III. Weight *W *of the ball.

IV. Reaction *R _{D} *at point

*D,*normal to horizontal surface.

The reactions *Rc *and *R*_{D}* *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 R _{A} and R_{C} 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 *R _{C} *acting at

*C*and normal to

*FD.*

*(iii) ** *Reaction *R*_{A}* *acting at *A *and normal to *DE. *

The reactions *R*_{A}* *and *R*_{C}* *will pass through *B, **i.e., *centre of the ball. The angles made by *R _{A} *and

*R*at point

_{C}*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 c**m 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 **F _{y}* = 0, we have

*R*–

_{C}*W*–

*F*sin θ = 0

*R _{C }*=

*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 R_{A} = Reaction at point A

*R*_{B}* *= Reaction at point *B *

*R _{C}* = Reaction at point C

The two rollers are also in contact at point *D. *Hence there will be a reaction R_{D}, 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 *R _{B} *at point

*B.*

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

*(iii) *Reaction R_{C} at point C. This will be normal to the vertical surface at point C

*(iv) *Reaction R_{D} at point D. This will be normal to the tangent at point D.

Fig. 27.7(b)

The reactions R_{B}, R_{C} and R_{D} 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 *R _{A} *at point A

*(iii)*Reaction

*R*at point D.

_{D}The reactions *R*_{A}* *and *R _{D} *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 R_{A} at point A

iii. Reaction R_{B} at point B

iv. Reaction R_{C} 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 R_{A} at point A. This is normal to OA.

*(iii) *Reaction R_{D} at point D. This is parallel to line OA.

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

For *F _{x} *= 0, we have

*R _{D} *sin 60°

*-R*sin 30° = 0 or

_{A}*R*sin 60° =

_{D}*R*sin 30°

_{A}

*R _{D} *cos 60° +

*R*cos 30° – 1000 = 0

_{A}*(0.577 R*cos 60° +

_{A})*R*cos30° = 1000 (

_{A}*R*= 0.577

_{D}*R*

_{A})or 0.577 × 0.5 *R A *+ *R _{A} *× 0.866 = 1000

*1.1545 R _{A} *= 1000 or

*R*= 1.1545 =

_{A}**866.17 N. Ans.**

Substituting this value in equation *(i), *we get

*R _{D} *= 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 *R _{B} *at point

*B*and normal to

*BO ;*

*(iii) *Reaction *R _{C} *at point C and normal to CO ; and

*(iu)*Reaction

*R*at point

_{D}*D*and parallel to

*BO.*

For *F _{x} *= 0, we have

*R _{B} *sin 30° +

*R*sin 60°

_{D}*– R*= 0

_{C}or *R _{B} *× 0.5 + 449.78 × 0.866

*– R*= 0

_{C}or *R _{C}*= 0.5

*RB*+ 432.8

*…*

*(ii)*

For *F _{y} *= 0, we have

*R _{B} *× cos 30° – 1000 –

*R*× cos 60° = 0

_{D}*or R _{B}* × 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 R_{C} = Reaction at C

and R_{D} = Reaction at D

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

The reaction R_{D} 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 *R _{D} *sin θ =

*R*

For= 0, we have

_{C}For

*R*cos θ = 1000

_{D}

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

Let *R _{A} *= Reaction at point

*A*

*R _{B} *= Reaction at point

*B*

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

The reactions *R _{A} R_{B} *and

*R*will pass through the centre

_{D}*F*of the sphere No. l.

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