Solved Problems in Engineering Mechanics

CHAPTER 1

Introductory Concepts

1.1 INTRODUCTION

Engineering Mechanics is the basic engineering discipline which reads with the bodies at motion or rest under the action of various forces. The scope of the subject starts from fundamental calculation methods of creating engineering design for under dynamic, atmospheric, temperature forces etc. The names of the scientists associated with the subject are Archemedes, Galilio, Newton, Einstein, Varignon, Euler, D Alembert etc and their contributions are discussed in the different chapters.

1.2 CLASSIFICATION OF ENGINEERING MECHANICS

Depending upon the nature of body under study engineering mechanics is divided into mechanics of rigid bodies which deals with rigid bodies and mechanics of fluids which deals with deformable bodies like fluid or gas (Fig.1.1).

Mechanics of Rigid bodies is classified into statics which deals with bodies at rest and dynamics which deals with bodies at motion.

Dynamics is further classified into Kinematics which deals with the motion of the body in term of its velocity, acceleration displacement maximum height reached without considering the forces that are responsible for its motion. The second one is Kinetics which deals with the motion of the body with the forces responsible.

In case of mechanics of deformable bodies the internal stresses are to be studied which are responsible for the deformation of the body.

1.3 BASIC TERMINOLOGY

Space, time mass and force all the important terms associated with mechanics in addition to length, displacement, velocity and acceleration.

Fig.1.1

1.4 LAWS OF MECHANICS

The subject is developed on the following fundamental laws.

1. Newton’s first law of motion

2. Newton’s second law of motion

3. Newton’s third law of motion

4. Newton’s law of gravitation

5. Law of transmissibility of forces

6. Parallelogram law of force

7. Varignon’s theorem

8. Polygon law of force

9. Lamis theorem

1.4.1 NEWTON’S FIRST LAWS OF MOTION

It states that unless external forces are applied a body which is at rest or in motion would continue to do so. The concept of inertia is developed from this law.

1.4.2 NEWTON’S SECOND LAW OF MOTION

According to this, the rate of change of momentum of a body is directly proportional of the applied force and body moves in the direction of force. From this the force is defined. The units of force in SI system is Newton

F = ma

1.4.3 NEWTON’S THIRD LAW OF MOTION

To every action there is equal and opposite reaction. This means that the forces of action and reaction between two bodies are equal in magnitude but opposite in direction.

1.4.4 NEWTON’S LAW OF GRAVITATION

According to this law, everybody attracts other body. The attractive force between two bodies is directly proportional to the mass of the bodies and inversely proportional to the square of the distance between them.

If two bodies of mass m1 and m2 are considered which are separated by a distance r, then the attractive force F is given by

Where G is called universal gravitational constant.

1.4.5 LAW OF TRANSMISSIBILITY OF FORCE

According to this law, the state of rest or motion of the body is not changed if a force acting on it is replaced by another of equal magnitude and direction but acting at any other point on the body.

1.4.6 PARALLELOGRAM LAW OF FORCES

This law enables us to determine the single force called resultant which can replace two forces acting on a point with the source effect caused by two forces. A states that if two forces acting simultaneously on a body at a point are represented in magnitude and direction by two adjacent sides of parallelogram, their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through the point of intersection of the two sides representing the forces.

Fig.1.2(a)

Fig.1.2(b)

As shown in Fig.1.2(a) at a point at in the body two forces F1 and F2 are acting R which is diagonal of the parallelogram ABCD constructed in such a way that AB represents magnitude and direction of F1 and AD represents magnitude and direction of F2 as shown in Fig.1.2(b).

Mathematically we can prove that

where θ is the angle between F1 and F2.

The direction of resultant R is given by α as shown in Fig.1.2(b) can be proved to be

1.4.7 VARIGNAN’S THEOREM

It states that the moment of a force about any point is equal to the sum of moments of the components of that force about the same point.

1.4.8 POLYGON LAW OF FORCE

It states that a number of explorer forces are acting at a point such that they can be represented in magnitude and direction by sides of polygon taken in order, their resultant is represented in both magnitude and direction by the closing side of the polygon taken in the opposite order.

Fig.1.3(a)

Fig.1.3(b)

As shown in Fig.1.3(a) four forces F1, F2, F3 and F4 are acting on point A. As shown in Fig.1.3(b) if all these forces are represented by sides of polygon both in magnitude and direction. The closing side AE gives the direction and magnitude of resultant R. Triangular Law of force which is widely used in mechanics is divided from polygon law of force.

1.4.9 LAMI’S THEOREM

If three concurrent forces P, Q and R acting on a body and keeping it in equilibrium then each force is proportional to the sine of the angle between other two forces.

Fig.1.4

As showing in Fig.1.4 three forces P, Q and R acting on a point A. The angle between the forces is given in the Fig.1.4.

According to Lami’s theorem

P α sin α, Q α sin β and R α sin γ

∴   We can write    

CHAPTER 2

System of Force

2.0 INTRODUCTION

The effect of a system of forces on a body is usually expressed in terms of its resultant which determines the motion of the body. Whenever this resultant is zero the body will be in equilibrium. In this chapter basic terms, fundamental principle related to Mechanics and techniques to evaluate resultant of forces are discussed.

System of Forces: When several forces of different magnitude and direction, act upon a body, they constitute system of Forces.

Coplanar Force System: If all the forces in a system lie on a single plane, it is called as coplanar force system.

Concurrent Force System: If the line of action of all forces pass through a same point, it is called as concurrent force system.

Collinear Force System: If the line of action of all forces lie along a single line then it is called a collinear force system. Various force system are shown in Fig.2.1

Fig.2.1

Rigid Body: A Rigid body is one in which relative position of any two points does not change under the action of forces.

Particle: A particle may be defined as an object which has only mass and no size; such a body cannot exist theoretically but when dealing with problems, involving distances considerably larger when compared to the size of the body.

Ex: Bomber aeroplane - Gunner operated from ground

A ship in the sea - control tower.

In study a movement of earth in celestial space, earth is treated as a particle.

Point Force: It is yet another assumption/idealization very commonly used in Engineering Mechanics. Ex: The weight of a man standing on an inclined ladder as shown in Fig.2.2. occupies some area of contact and cannot apply his wt through a single point. However this area is small compared to other dimensions. Hence not much accuracy is lost by treating it as point force and thereby simplifying the problem.

Fig.2.2

2.1 RESOLUTION OF FORCES

Is exactly opposite process of composition of forces It is the process of finding the number of components which will have the same effect on the body as given by single force

Fig.2.3(a)

Problem 2.1: Fig.2.3(a) shows the force P. Find horizontal and vertical components

Sol: Py = P cos 30 = 20 cos 30 = 17.321 kN (Down)

Px = P sin 30 = 20 sin 30 = 10 kN (left)

Problem 2.2: As shown in Fig.2.3(b), a load W is placed on the incline. Find its components

Sol: W = 10 kN

PN = W cos 20 = 10 cos 20 = 9.3969 kN

Fig.2.3(b)

Fig.2.3(c)

PH = Wsin 20° = 10 sin 20 = 3.4202 kN

2.3 COMPOSITION OF FORCES BY METHOD OF RESOLUTION

Whenever multiple forces are acting on a body as shown in Fig. 2.3(c), the resultant of all the forces is found. First, the component of each force along X and Y direction is determined and added algebraically to get two components. These two are combined to get resultant. The X and Y components of P1, P2 P3 and P4 are evaluated and all x components and x components are added algebraically as given in the equation Net resultant will be

Force component in ΣX = P1x + P2x + P3x + P4x

ΣY = P1y + P2y + P3y + P4y

Problem 2.3: As shown in Fig. 2.4, if θ = 60, W = 1000 N, F = 100 N, T = 1200 N and N = 500 N. Find the resultant

Fig.2.4

Sol: Assuming X-axis along the plane

                 Y-axis perpendicular to the plane

ΣX = 0 ⇒ F – W sin θ + T = 0

Or      ΣX = F – W sin θ – T = 0 (+ve, downwards)

       = 100 - 1000 sin 60 – 1200

       = 433.97 acts upwards

ΣY = N – W cos θ

       = 500 – 1000 cos 60 = 0

Vertical forces are balanced

Net resultant upwards on the plane = 433.97 N

Problem 2.4: Two forces acting on a body are 500N and 1000N as shown in Fig.2.5. Determine the third force P such that the resultant of all the three forces is 1000N directed at 45° to X-axis.

Fig.2.5

Sol: Let the third force if P acting at θ to X-axis as shown in Fig.2.5

Applying ΣX = 0

1000 cos 45 = 500 cos 30° + P cos θ + 1000 cos 60° ⇒ P cos θ = –255.9

Applying ΣY = 0

1000 sin 45 = 500 sin 30° + P sin θ + 1000 sin 60° ⇒ P sin θ = –408.9

Fig.2.5(a)

Problem 2.5: Three forces acting at appoint are shown in Fig.2.6. The direction of 300 N forces may vary, but the angle between the is always 40°. Determine the value of θ for which the resultant of the three forces is parallel to plane.

Fig.2.6

Sol: Assume local planes as shown in Fig.2.6.

Resultant to act along the plane

∴  ΣY = 0

300 sin θ + 300 sin(40 + θ) – 500 sin 30 = 0

⇒      sin θ + sin(40 + θ) = sin 30 =

[sin C + sin D formula]

sin(40 + θ) + sin θ =

2 sin (20 + θ) cos 20 =

∴      θ = 6.35°

Problem 2.6: Determine the horizontal force P to be applied to a block shown in Fig.2.7(a) of weight 1200 N to hold it in position on a sooth inclined plane AB which makes an angle 30° with the horizontal

Sol: The free body diagram is shown in Fig.2.7(b), R is the normal reaction

Fig.2.7(a)

Fig.2.7(b)

Applying ΣY = –1200 + R sin 60 = 0

R sin 60° = + 1200

R = 1385.6 N

Applying ΣX = 0

R cos 60° = +P

R sin 60° = + 1200

tan 60° =

∴ P = 1200/tan 60° = 692.8 N

Problem 2.7: A roller of 10 kN rests on a smooth horizontal floor and is connected to the floor by the bar AC as shown in Fig.2.8(a). Determine the force in the bar AC and reaction from floor, if the roller is subjected to a horizontal force of 5kN and an inclined force of 7 kN as shown in Fig.2.8(a).

Sol: The force developed in AC may be Tensile or Compressive. Assuming compressive force. Free body diagram is shown in Fig.2.8(a).

Fig. 2.8(a)

Fig. 2.8(b)

Fig. 2.8(c)

ΣX = 5 – 7 cos 45° + S cos 30 = 0

∴ S = –0.058 kN

ΣY = R + S sin 30 – 10 – 7 sin 45 = 0

Problem 2.8: A system of connected flexible cables shown in Fig.2.9(a) is supporting two vertical forces 150 N and 200 N at points B and D. Determine the forces in various segments of the cable.

Sol: Free body diagrams of the points B and D are shown in Fig.2.9(a) and (b) respectively.

Tension in cable DE = T1

Tension in cable BD = T2

Tension in cable BC = T3

Tension in cable AB = T4

Fig.2.9(a)

Fig.2.9(b)

Fig.2.9(c)

Applying Lami’s theorem for Point D Fig.2.9(b)

∴ T1 = 179.32 N and T2 = 146.4 N

From Fig.2.9(c)

ΣY = 150 – T3 cos 30 + T2 cos 60 = 0

ΣH = T3 sin 30 – T4 + T2 sin 60 = 0

T4 = T3 sin 30 + T2 sin 60 = 255.65 N

Problem 2.9: A rope AB, 4.5 long is connected at two points A and B at the same level 4 m apart. A load of 1500 N is suspended from a point C on the rope 1.5 m from A as shown in Fig.2.10(a). What load connected at point D on the rope from B will be necessary to keep the position CD level AB = 4.0 m

Sol: As shown in Fig.2.10(a) CE and DF are drawn perpendicular to AB.

Fig.2.10(a)

Fig.2.10(b)

Fig.2.10(c)

If CE = y and AE = x

In Triangle AEC

x² + y² = 1.5²

BF = AB – (AE+EF) = 4–(x + 2) = 2 – x

Also, BF² + DF² = 1²

(2.0 – x)² + y² = 1      …..(2)

4 + x² – 4x + y² = 1

4 + 2.25 – 4x = 1

4x = 5.25

x = 1.3125

From ACE,         

α = 28.955° = 29

Let Tension in AC = T1 and

Tension in CD = T2

T1 = 3098.4 N         T2 = 2711.1 N

Applying Lami’s theorem for forces at B

T3 = 3098.39 N

W = 2711.1 N

Problem 2.10: A rigid prismatic weightless bar AB is supported in a vertical plane by a hinge at the end A and by a horizontal string attached to the bar at D as shown in Fig.2.11(a). The end B of the bar carries a load W. Find the tension in the string and the direction of the reaction at the hinge in terms of W and θ.

Sol: Free body diagram is shown in Fig.2.11(b)

Fig.2.11(a)

Fig.2.11(b)

Let the reaction RA make an angle α with the horizontal as shown in Fig.2.11(b)

In the Δ BED

In the ΔBFA

From (1) and (2)

Since E is the Midpoint of the BF

In the ΔBAF

AF = lsinθ [∴ AB = l]

Dividing (3) by (4) we get (or)

Consider the equilibrium of the bar AB

Problem 2.11: A uniform wheel of 60cm diameter weight 1000 N just against a rectangular block 15 cm height lying on a horizontal plane as shown. Fig.2.12(a) It is to be pulled over this block by a horizontal force P applied to the end of a string wound round the circumference of the wheel. Find the force P when the wheel is just about to roll over the block.

Sol: Free body diagram is shown in Fig.2.12(b). Let RA is the reaction at A

AD² = AC² – CD²

        = 30² – 15²

∴     AD = 25.98 cm

Fig.2.12(a)

Fig.2.12(b)

Now from the Fig. BD = 15 cm; CD = 15 cm; AD = 25.98 cm

the ΔADE,

θ = 29.98 ≅ 30°

Writing the equation of equilibrium

ΣX = 0    ⇒    P – RA sin θ = 0  ⇒  P = RA sinθ

ΣY = 0    ⇒  –W + RA cos θ = 0  ⇒  W = RA cosθ

Problem 2.12: Two forces equal to 2P and P respectively act on a particle. If first be doubled and the second increased by 12 N the direction of the resultant is unaltered, Find the value of ‘P’.

Sol: Let the inclination of resultant be θ, in the first and second case, we have, (Using parallelogram law of forces)

CASE 1:

CASE 2:

Given that the two forces are equal so, equating land 2

4P + (P + 12) cos α = (P + 12) (2 + cos α)

    = 2P + 24 + P cos α + 12 cos α

4P + (P + 12) cosα = 2P + 24 + (P + 12) cos α

4P = 2P + 24

2P = 24

P = 12 N

Problem 2.13: Find the Reactions RA and RB included at the supports A and B of the right angle bar ACB supported as shown in Fig.2.13(a) and subjected to a vertical load P applied at the Midpoint of AC.

Sol: Free body diagram is shown in Fig.2.13(a). The reaction are also shown in Fig.2.13(b).

Fig.2.13(a)

Fig.2.13(b)

Under equilibrium condition

ΣX = 0   ⇒   RB – Ax = 0   ⇒   RB = Ax        …..(1)

ΣY = 0   ⇒   RA – P = 0   ⇒ RA = P        …..(2)

Consider Moment of all forces about ‘A’

ΣMA = 0

0.6 – 0.9 RB = 0

RB = P = P

From eq.(i)

Ax = RB = P

RA = P and RB = P

Problem 2.14: Three identical cylinders each weighing W, are stacked as shown in Fig.2.14(a) below, an smooth inclined surfaces each inclined at an angle ‘θ’ with the horizontal. Determine the small angle ‘θ’ to prevent stack from collapsing.

Sol: Free body diagram is shown in Fig.2.14(b)

Fig.2.14(a)

Fig.2.14(b)

Fig.2.14(c)

Now from Lami’s theorem

ΔABC are equilateral triangle

R1 = R2 = 0.577 W

Free body diagram of cylinder C2 is shown in Fig.2.14(d).

Applying equilibrium condition for A

Fig.2.14(d)

ΣX = 0

–R3 – R1 cos 60 + R4 cos(90 – θ) = 0

R4 sin θ = R3 + R1 cos 60

Here R3 = 0 because while their collapsing they lose contact

ΣY = 0 ⇒ R4 sin(90 – θ) – R1 sin 60 – W = 0

0.2885 cot – = 1 + (0.577 B/2)

0.2885 cot θ= 1.499

Problem 2.15: A uniform rod AB of negligible weight is hinged at A and supported at B by a string as shown in Fig.2.15(a). Find the value of angle θ corresponding position of equilibrium of the bar if Q = P/2.

Sol: Free body diagram is shown in Fig.2.15(b).

Fig.2.15(a)

Fig.2.15(b)

Fig.2.15(c)

2α = 180 – θ

α = 90 – α/2

Now applying ΣMA = 0

P × l cos θ – Q × l cos θ/2 = 0

2PK – QK – P = 0

Problem 2.16: A body acted upon by 3 forces F1, F2 and F3 is in equilibrium is shown in Fig.2.16(a). If the magnitude of force F3 is 500 N. Find forces F1 and F2.

Fig.2.16(a)

Fig.2.16(b)

Sol: From Fig.2.16

ΣH = 0, Horizontal forces

– F1 cos 50 + F2 cos 30 = 0          …..(1)

ΣV = 0

–500 + F2 sin 30 + F1 sin 50 = 0          …..(2)

F2 cos 30 = F1 cos 50

F1 = F2 (1.35)          …..(3)

F2 sin 30 + F2 (1.35) sin 50 – 500 = 0

F2 (05) + F2 × 1.355 × 0.766 – 500 = 0

F2 = 326.8 N

F1 = 1.35 × F2

F1 = 441.2 N

Another Method

F2 = 326.4 N (Lami’s theorem can also be used)

Problem 2.17: A force P = 500 N is applied at the centre C of beam AB of length 5m as shown in Fig.2.17(a). Find the reaction at hinged and roller support.

Sol: Free body diagram is shown in Fig.2.17(b)

Fig.2.17(a)

Fig.2.17(b)

ΔBCD, From Fig. 2.17(a)

BD = CB tan 30°

      =