# Free Body Diagram

FREE BODY DIAGRAM

To analyse the forces acting on a body, the body is to be mechanically isolated from all surrounding bodies. Only the forces acting on the body by the surrounding bodies are taken into consideration. A diagram consisting of the isolated body with all external forces on it, is called a free body diagram. There are various steps followed in drawing free body diagram.

1. Depending upon the problem, the free body is selected. This body is then isolated from the ground and is separated from surrounding bodies. The contour of the free body is sketched.
2.  On the free body, all the external forces exerted on the free body by the ground and by the  detached bodies are to be considered. The weight of the free body is  also to be taken into account. The weight acts at the center of gravity of the body
3.  The forces shown on the free-body diagram must be those which are exerted on, and not by, the free body.
4.  Support reactions on the free body should be indicated properly.
5.  There should be all dimensions in the free body diagram because these are required to calculate moments of the forces. We c.an omit other details.

Problem 2.9. Two smooth circular cylinders, each of weight W = 1000 N and radius 15 cm are connected at their centres by a string AB of length = 40 cm and rest upon a horizontal plane, supporting above them a third cylinder of weight = 2000 N and radius 15 cm as shown in Fig.2.10. Find the force S in the string AB and the pressure produced on the floor at the points of contact D and E.

Sol. Given:

Weight of cylinders 1 and 2 = 1000 N

Weight of cylinder 3 = 2000 N
Radius of each cylinder = 15 cm

Length of string AB = 40 cm

From Fig. 2.10 (b), AC = AF + FC = 15 + 15 = 30 cm

Equilibrium of cylinder 3.The cylinder 3 has points of contact at F and G. The reactions RF and Ra will pass through the centre of sphere 3. The free-body diagram is shown in Fig. 2.10(c). Resolving forces horizontally,

RF sin θ – RG sin θ = 0
or RF = RG … (i)
Resolving forces vertically,
RF cos θ + RG cos θ = 2000
or RF cos θ + RG cos θ = 2000 RF = RG)

Equilibrium of cylinder 1. The cylinder 1 has points of contact at D and F. Also the cylinder 1 is connected to cylinder 2 by a string AB. To draw the free-body diagram of cylinder 1, there will be reactions RF and RD at points F and D as shown in Fig.2.9 (d). Also there will be a force S in the direction of the string AB.

For Fx= 0, we have
S -RF sin θ = 0

S = RF sin θ

= 1342.179 x sin 41.836°

[ From equation (ii), RF = 1342.179)
= 895.2 N. Ans.

For Fy= 0, we have

RD – 1000 – RF cos θ = 0

RD = 1000 + RF cos θ

= 1000 + 1342.179 × cos 41.836

= 1999.99 = 2000 N. Ans.

The equilibrium of cylinders 1, 2 and 3 taken together.The three cylinders taken together have points of contact at D and E. The free-body diagram is shown in Fig. 2.9 (e). In this case only vertical forces exist. Hence resultant force in y-direction should be zero.

RD + RE – 1000 – 2000 -1000 = 0

or   RE = 1000 + 2000 + 1000 – RD
RE
= 4000 – RD

= 4000 – 2000                                          ( RD = 2000)

=2000 N. Ans.

Fig. 2.10(e)

Problem 2.10.A roller of radius 40 cm, weighing 3000 N is to be pulled over a rectangular block of height 20 cm as shown in Fig. 2.11, by a horizontal force applied at the end of a string wound round the circumference of the roller. Find the magnitude of the horizontal force which will just turn the roller over the corner of the rectangular block. Also determine the magnitude and direction of reactions at A and B. All surfaces may be taken as smooth.

Sol. Given:

Radius of roller = 40 cm

Weight, W = 3000 N

Height of block = 20 cm

Find horizontal force P, reaction RA and reaction RB when the roller just turns over the block.

When the roller is about to turn over the corner of the rectangular block, the roller lifts at the point A and then there will be no contact between the roller and the point A. Hence reaction RA at point A will become zero.

Now the roller will be in equilibrium under the notion of the following three forces:

i.            Its weight W acting vertically downward

ii.            Horizontal force P

iii.            Reaction RB at point B. The direction of RB is unknown.

For the equilibrium, these three forces should pass through a common point. As the force P and weight W is passing through point C, hence the reaction RB must also pass through the point C. Therefore, the line BC gives the direction of the reaction RB.

2nd Method

This problem can also be solved by taking moments of all the three forces about the point B (i.e., corner of the rectangular block) as shown below:

P × CD = W × BD

or                                 P × 60 = 3000 × 34.64                                         (   BD = 34.64)

Problem 2.11. If in the problem 2.10, the force P is applied horizontally at the centre of the roller, what would be the magnitude of this force? Also determine the least force and its line of action at the roller centre, for turning the roller over the rectangular block.

So1. Given:

Radius of roller = 40 cm

Weight, W = 3000 N

Height of block = 20 cm

When the roller is about to turn over the corner of the rectangular block, the roller lifts at the point A, and then there will be no contact between the roller and the point A. Hence reaction RA at point A will be zero

Now the roller will be in equilibrium under the action of the following three forces:

i.            Weight of the roller W acting vertically downward through point O.

ii.            Horizontal force P applied at the centre of the roller

iii.            Reaction RB at point B. The direction of reaction RB is unknown.

The above three forces should pass through a common point, as roller is in equilibrium. But the weight W and force P is passing through point O, hence RB should also pass through point O. Hence line joining B to O, gives the direction of RB as shown in Fig. 2.12.

Method of moments

The force P can also be calculated by the method of moments. Taking the moments of all the forces about the point B, we get

P × OD – W × BD = 0

Least force and its line of action

Let Pmin = Least force applied as shown in Fig. 2.13.
α = Angle of the least force

From OBC, BC = BO sin α

Taking moments of all forces about point B, we get Pmin × BC – W × BD = 0

The force P will be minimum when sin α is maximum. But sin α will be maximum, when α = 90° or sin α = 1. Substituting this value of sin α in the above equation, we get minimum force.

The direction of least force i.e., Pmin) is at right angles to the line BO.

Problem 2.12. A L-shaped body ABC is hinged at A with a force F acting at its end C. Determine the angle θ which this force should make with the horizontal to keep the edge AB of the body vertical.

Sol. Given:

Length BC = 25 cm

Length AB = 40 cm

Let us assume that force F is making an angle θ with the horizontal as shown in Fig. 2.14.

The forces acting on the given body ABC are:

(i) Force F acting at point C.

(ii) Reaction RA at a point A.

Only two forces are acting on the body and the body is in equilibrium. But the two forces can be in equilibrium only if they are equal in magnitude, opposite in direction and have the same line of action. As the force F is acting at point C, hence reaction RA should also pass through point C. Therefore, the line of action of reaction RA is along line AC. The line of action of F should also be along AC, i.e., the force F will be making an angle α with the horizontal.

Problem 2.13.A horizontal force 200 N is applied to the sloping bar BCD whose bottom rests on a horizontal plane, as shown in Fig. 2.15. Its upper end is pinned at B to the horizontal bar AB which has a pinned support at A. What couple M must be applied to AB to hold the system in equilibrium? What is the magnitude of the pin reaction at B? Assume the bars to be weightless and pins at A and B to be smooth.

Sol. Given:

Length AB = 1.6 m, Length BD = 1.2 m, BC = 0.8 m and CD = 0.4 m

Horizontal force at C = 200 N

Let M = Couple applied A to bar AB to keep the system in equilibrium.

RB = Reaction at B

RBX and RBY are the horizontal and vertical components of reaction RB.

The free-body diagram of the sloping bar BCD is shown in Fig. 2.16 (b). The reaction RD at point D is vertical. Also show RBX and RBY at B.

Since the bar is in equilibrium, first apply Fx= 0, which gives

RBX= 200 N

Now apply Fy= 0, then RBY  = RD                                            … (i)
The magnitude of RBY and RD is unknown. To find their values apply M at the pin B.
Taking moments of all forces at point B, we get

RD × DD’ = 200 × BC’

or RD × BD cos 60° = 200 × BC sin 60 ( DD’ =BD cos 60° andBC’ = BC sin 60°)

or RD × 1.2 cos 60° = 200 × 0.8 sin 60°

Now draw the free-body diagram of the horizontal bar AB as shown in Fig. 2.16 (a) how the reactions RBX and RBY at B. Also show the reactions RAY and RAX at A. The applied couple M is also shown in the figure.

Apply MA = 0

M = RBY × 1.6 = 230.93 × 1.6
= 369.44 Nm. Ans.

Problem 2.14. A body weighing 2000 N is suspended with a chain AB 2 m long. It is pulled by a horizontal force of 320 N as shown in Fig. 2.17. Find the force in the chain and the lateral displacement (i.e., x) of the body.

Sol. Given:

Weight suspended at B = 2000 N

Length AB = 2 m

Horizontal force at B = 320 N

Find : Force in AB and value of x

Let F = Force in chain AB

θ = Angle made by AB with horizontal.

The free body diagram of the point B is shown in Fig 2.17 (b)

The point B is in equilibrium under the action of three forces. Hence using Lami’s theorem, we get