UNIT2
EQUILIBRIUM OF FORCES
Equilibrium of force
To apply the equation of equilibrium, we have to consider all types of forces (ΣF) which are acting on the particle. A drawing in which all the forces are depicted is called free body diagram (FBD).
Process to Draw a FreeBody Diagram
To construct a free body diagram, the following three steps are necessary.
Draw Outlined Shape.
Imagine that the particle is to be isolated from its environments by drawing its outlined shape.
Show all the forces.
Then on this sketch, show all the types of forces. Identify Each Force that are known and unknown and display them.
Identify each force
Known forces are displayed with magnitude and direction. Letters are used to represent the magnitudes and directions of forces that are unknown.
For understanding the freebody diagram, we will consider the following example:
Fig 1
Suppose, in this example there are two strings (AC and BC) attached to an object at point C. Thus, the first step involves drawing a simple diagram representing the outline of all components.
Then, we need to show all the forces that are contributing in the given diagram. Here, there is 15 N force acting at point C, and there is tension in strings BC and AC due to 15 N force acting at C. Thus, identifying and displaying these forces onto the outline diagram is the second step.
The third step is finding the magnitude and direction of the forces and displaying them on the outline diagram. Using these steps, we can draw the freebody diagram as shown below.
Fig 2
If three coplanar concurrent forces are in equilibrium then, the ratio of magnitude of any force to the sine of angle between the other two is constant.

Fig 3
According to Lami’s Theorem:
Lami’s theorem can also be written as: If three coplanar concurrent forces acting at a point are in equilibrium, each force is proportional to the sine of angle between other two.
Key takeaways:
Lami’s Theorem:
Lami’s theorem can also be written as: If three coplanar concurrent forces acting at a point are in equilibrium, each force is proportional to the sine of angle between other two.
1. If the body moves in any direction, it means that there is resultant force acting on it. In other words, the horizontal component of all the forces (Σ H) and vertical component of all the forces (ΣV) must be zero. Mathematically,
Σ H = 0 and Σ V = 0
2. If there is rotation of body, without moving, that means there is couple acting. In other words, the resultant moment of all the forces (Σ M) must be zero. Mathematically,
Σ M = 0
3. If the body moves in any direction and at the same time it rotates about itself, if means that there is a resultant force and also a resultant couple acting on it. In other words, horizontal component of all the forces (Σ H), vertical component of all the forces (Σ V) and resultant moment of all the forces (Σ M) must be zero. Mathematically,
Σ H = 0 Σ V = 0 and Σ M = 0
4. If the body is completely at rest it means there is neither force nor any couple acting on it. A little consideration will show that in this case the following conditions are already satisfied:
Σ H = 0 Σ V = 0 and Σ M = 0
The abovementioned equations are known as the conditions/equations of equilibrium.
Thus, the equations of equilibrium for twodimensional coplanar force system are given by considering that the particle is present in xy plane as shown in fig. below, then the force can be resolved into i and j components.
Fig 4
For equilibrium, the forces must add on to produce a zeroforce output, i.e.,
Σ F = 0
Σ Fx i + Σ Fy j = 0
Thus, for satisfying this vector equation, the x and y component must be zero as mentioned in earlier points. Hence,
Σ Fx = 0
Σ Fy = 0
Thus, for solving twodimensional coplanar force problems using conditions of equilibrium, we can follow the belowmentioned procedure.
FreeBody Diagram.
• In any orientation, establish x and y axes.
• Then mark all the known and unknown forces in figure.
• Unknown forces are also assumed and plotted.
Equilibrium equation.
• Apply the equations of equilibrium, Σ Fx = 0 andΣ Fy = 0.
• The forces are directed as positive for those along positive axis and viceversa.
• If the resultant comes out to be negative value, it means that the direction needs to be reversed for that is already considered.
As we know, the sufficient case for stating equilibrium for a body under combination of forces isΣ F = 0.
For any threedimensional force system, as shown in Fig. below, we can resolve the forces into their respective i, j, k components, so that
Σ Fx i + Σ Fy j + Σ Fz k = 0
Fig 5
To satisfy this equation we require,
Σ Fx = 0 ,Σ Fy = 0 , Σ Fz = 0
These three equations state that the algebraic sum of the components of all the forces acting on the particle along each of the coordinate axes must be zero. Using these equations mentioned we can solve for maximum three of the unknowns, shown by coordinate direction angles or magnitudes of forces shown on the particle’s freebody diagram.
For solving three dimensional problems we can use the following procedure.
Free Body Diagram:
• In any orientation, establish x and y axes.
• Then mark all the known and unknown forces in figure.
• Unknown forces are also assumed and plotted.
Equations of Equilibrium.
• Apply the equations of equilibrium, Σ Fx = 0, Σ Fy = 0 andΣ Fz = 0.
• Initially we can display as a Cartesian vector if it becomes to solve the problem regularly, substitute these vectors into Σ F = 0 and then set the i j, k components equal to zero.
• If the resultant comes out to be negative value, it means that the direction needs to be reversed for that is already considered.
Key takeaways:
Three equations state that the algebraic sum of the components of all the forces acting on the particle along each of the coordinate axes must be zero
Coplanar concurrent forces
The sum of all forces in the xdirection or horizontal is zero.
ΣFx=0 or ΣFH=0
The sum of all forces in the ydirection or vertical is zero.
ΣFy=0 or ΣFV=0
Important Points for Equilibrium Forces
Two forces are in equilibrium if they are equal and oppositely directed.
Three coplanar forces in equilibrium are concurrent.
Three or more concurrent forces in equilibrium form a close polygon when connected in headtotail manner.
Equilibrium of Coplanar concurrent forces
Concurrent means that the forces intersect through a single point.If forces are concurrent, we can add them together as vectors to get the resultant.
If the body is not accelerating, it must be in equilibrium, so that means the resultant is zero
For concurrent forces, the body is a point. So for concurrent forces in equilibrium, the forces should all add up to give zero.
If a body is not accelerating is in equilibrium, so resultant of all forces = 0.
A typical concurrent force situation is a lifting eye. The pulling forces in any cables must pass through the centre of the eye. If there is only one eyebolt (correctly positioned over the centre of gravity) and the load is suspended, the bolt force must pass through the same centre. Hence all forces pass through one point (the centre of the eye), so we have concurrent equilibrium.
Coplanar nonconcurrent forces
It is defined as the forces which lies in the line of activity but do not intersect or forms a parallel at a common point. The forces are the general system of forces.
Show the nonconcurrent, nonparallel systems in figure (1).
Coplanar = within the same plane.
Concurrent = passing through the same point.
Fig 6
In figure ( ), there are four forces; F1, F2, F3 and F4 respectively. These forces act on a single A plane.
Consider an example for nonconcurrent, nonparallel systems. A man is standing on the ladder which is resting against a wall. Here, the forces on the ladder do not meet at a common point but it lies on a single plane which is the wall.
The resultant of nonconcurrent, nonparallel systems is obtained by the principles of equilibrium. In nonconcurrent force system, consider three equilibrium conditions; force equilibriums in xdirection and y&direction, and moment equilibrium by the sum of moments about point O.

The pair of force system is solved graphically along with parallelogram or triangle law. After calculating the pair of forces, calculate the resultant of the forces and combine it to determine the new resultant.
Equilibrium of Coplanar nonconcurrent forces
There are three equilibrium conditions that can be used for nonconcurrent, nonparallel force system.
The sum of all forces in the xdirection or horizontal is zero.
ΣFx=0or ΣFH=0
The sum of all forces in the ydirection or vertical is zero.
ΣFy=0 or ΣFV=0
The sum of moment at any point O is zero.
ΣMO=0
The three equilibrium conditions can solved up to three unknowns in the system. If the system involves more than three unknowns, it is called indeterminate.
Key takeaways:
Coplanar = within the same plane.
Concurrent = passing through the same point.
If a body is not accelerating is in equilibrium, so resultant of all forces = 0.
Friction
Friction is the force that opposes motion between any surfaces that are in contact. There are four types of friction: static, sliding, rolling, and fluid friction. Static, sliding, and rolling friction occur between solid surfaces. Fluid friction occurs in liquids and gases. All four types of friction are described below.
Static Friction
Static friction acts on objects when they are resting on a surface. For example, if you are hiking in the woods, there is static friction between your shoes and the trail each time you put down your foot (see Figure below). Without this static friction, your feet would slip out from under you, making it difficult to walk. In fact, that's exactly what happens if you try to walk on ice. That's because ice is very slippery and offers very little friction.
Sliding Friction
Sliding friction is friction that acts on objects when they are sliding over a surface. Sliding friction is weaker than static friction. That's why it's easier to slide a piece of furniture over the floor after you start it moving than it is to get it moving in the first place. Sliding friction can be useful. For example, you use sliding friction when you write with a pencil. The pencil “lead” slides easily over the paper, but there's just enough friction between the pencil and paper to leave a mark.
Rolling Friction
Rolling friction is friction that acts on objects when they are rolling over a surface. Rolling friction is much weaker than sliding friction or static friction. This explains why most forms of ground transportation use wheels, including bicycles, cars, 4wheelers, roller skates, scooters, and skateboards. Ball bearings are another use of rolling friction. You can see what they look like in the Figure below. They let parts of a wheel or other machine roll rather than slide over on another.
Fig 7
The ball bearings in this wheel reduce friction between the inner and outer cylinders when they turn.[Figure 7]
Fluid Friction
Fluid friction is friction that acts on objects that are moving through a fluid. A fluid is a substance that can flow and take the shape of its container. Fluids include liquids and gases. If you've ever tried to push your open hand through the water in a tub or pool, then you've experienced fluid friction. You can feel the resistance of the water against your hand. Look at the skydiver in the Figure below. He's falling toward Earth with a parachute. Resistance of the air against the parachute slows his descent. The faster or larger a moving object is, the greater is the fluid friction resisting its motion. That's why there is greater air resistance against the parachute than the skydiver's body.
Dry friction resists relative lateral motion of two solid surfaces in contact. The two regimes of dry friction are 'static friction' ("stiction") between nonmoving surfaces, and kinetic friction (sometimes called sliding friction or dynamic friction) between moving surfaces.
Ff ≤μFn
Where
Ff is the force of friction exerted by each surface on the other. It is parallel to the surface, in a direction opposite to the net applied force.
μ is the coefficient of friction, which is an empirical property of the contacting materials,
Fn is the normal force exerted by each surface on the other, directed perpendicular (normal) to the surface.
The Coulomb friction Ff may take any value from zero up to μFn , and the direction of the frictional force against a surface is opposite to the motion that surface would experience in the absence of friction. Thus, in the static case, the frictional force is exactly what it must be in order to prevent motion between the surfaces; it balances the net force tending to cause such motion. In this case, rather than providing an estimate of the actual frictional force, the Coulomb approximation provides a threshold value for this force, above which motion would commence. This maximum force is known as traction.
The force of friction is always exerted in a direction that opposes movement (for kinetic friction) or potential movement (for static friction) between the two surfaces. For example, a curling stone sliding along the ice experiences a kinetic force slowing it down. For an example of potential movement, the drive wheels of an accelerating car experience a frictional force pointing forward; if they did not, the wheels would spin, and the rubber would slide backwards along the pavement. Note that it is not the direction of movement of the vehicle they oppose it is the direction of (potential) sliding between tire and road.
It is the highest value of static friction which comes into play when an object is just about to slide over the surface of a different object. For an exerted external force greater than the limiting friction, the body begins to move.
Once the motion has started, static friction cannot be considered. A new type of friction termed as kinetic friction comes into play.
The direction of limiting frictional force is always contrary to the direction of motion.
Limiting friction acts tangentially to the two surfaces interacting.
The magnitude of limiting friction is proportional to the normal reaction amidst the two surfaces.
The limiting friction hinges upon the material, the nature of the surfaces interacting, and their evenness.
So long as the normal reaction is the same, the magnitude of limiting friction is free of the shape or the area of the surfaces in contact, for any two given surfaces.
Laws of Static Friction:
Following are the laws of static friction:
1. The force of friction always acts in direction which is opposite to motion of body.
2. The magnitude of force of friction is equal to the force acting on the body externally.
3. There is constant ratio between limiting frictional force and the normal reaction forces. Mathematically:
(F/R) = Constant
Where F = Limiting friction, and R = Normal reaction.
4. The force of friction does not depend on the area of contact between the two surfaces.
5. There is dependency of frictional force on roughness of the surfaces.
Laws of Kinematic or Dynamic Friction:
Following are the laws of kinetic or dynamic friction:
1. The force of friction always act such that it opposes the motion of body.
2. There is a constant ratio between kinetic motion and normal reaction. But this ratio is slightly less when it comes to limiting friction.
3. For medium speeds, the force due to friction stays constant. But it decreases somewhat when the speed increases.
e.g. 1: The uniform crate shown in Fig 9. below has a mass of 20 kg. If a force P = 80 N is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is = 0.3.
Fig 9
Solution:
The first step is freebody diagram i.e.
FreeBody Diagram. As shown in Fig. diagram below, the resultant normal force NC acts at a distance of x from the Centre line of crate for countering the tilting effect caused by P. Hence, there are three unknowns, F, NC and x, which can be solved by equations of equilibrium.
Fig 10
Equations of equilibrium:
Σ Fx = 0
80 cos 30° N  F = 0
Σ Fy = 0
80 sin 30° N + NC  196.2 N = 0
Σ M= 0
80 sin 30° N10.4 m2  80 cos 30° N10.2 m2 + NC (x) = 0
Thus, solving the above equations, we get,
F = 69.3 N
NC = 236 N
x = 0.00908 m = 9.08 mm
As value of x is negative, hence it acts to left of centre line of the crate. There will be no tipping since x < 0.4 m. Fmax = Nc = 0.3(236 N) = 70.8 N. Since F that we got is 69.3 N < 70.8 N, the crate will not slip.
E.g. 1: A body of weight 300 N is lying on a rough horizontal plane having a coefficient of friction as 0.3. Determine the magnitude of the force, which is needed to move the body, acting at 25° by the horizontal.
Fig 11
Solution:
Given:
Weight of the body (W) = 300 N; Coefficient of friction (μ) = 0.3 and
angle made by the force with the horizontal (α) = 25°
Let P = Magnitude of the force, which can move the body, and
F = Force of friction.
Initially, we will solve the problem in horizontal direction as,
The next step is to resolve the force in vertical direction, we get,
The force of friction (F) is given as,
Hence, P = 87.1 N.
E.g. 3: A body, which is positioned on a rough horizontal plane, needs a pull of 180 N inclined at 30° to the plane just for moving. Also, it was found that a push of 220 N which is inclined at 30° to the plane also moved the body. Find out the weight of the body and the coefficient of friction present here.

Fig 12
Solution:
Given: Pull = 180 N; Push = 220 N and angle at which force is inclined with horizontal plane (α) = 30°
Let W = Weight of the body
R = Normal reaction, and
μ = Coefficient of friction.
First of all, consider a pull of 180 N acting on the body. We know that in this case, the force of friction (F1) will act towards left as shown in Fig. (a).
Resolving the forces horizontally,
F1 = 180 cos 30° = 180 × 0.866 = 155.9 N
And now resolving the forces vertically,
R1 = W – 180 sin 30° = W – 180 × 0.5 = W – 90 N
We know that the force of friction (F1),
155.9 = μR1 = μ (W – 90)
Now consider a push of 220 N acting on the body. We know that in this case, the force of friction (F2) will act towards right as shown in Fig. (b).
Resolving the forces horizontally,
F2 = 220 cos 30° = 220 × 0.866 = 190.5 N
And now resolving the forces horizontally,
R2 = W + 220 sin 30° = W + 220 × 0.5 = W + 110 N
We know that the force of friction (F2),
190.5 = μ.R2 = μ (W + 110) ..(ii)
Dividing equation (i) by (ii)
155.9 W + 17 149 = 190.5 W – 17 145
34.6 W = 34 294
W = 991.2 N
Now substituting the value of W in equation (i),
155.9 = μ (991.2 – 90) = 901.2 μ
μ = 0.173
In this way, numericals based on dry friction can be solved.
A wedge is a piece of metal or wood which is usually of a triangular or trapezoidal in crosssection. It is used for either lifting loads through small vertical distances or used for slight adjustments in the position of a body i.e., for tightening fits or keys for shafts.
When lifting a heavy load the wedge is placed below the load and a horizontal force P is applied as shown in Fig. 3.41. If the force P is just sufficient to lift the load, the wedge will move towards left and load will move up. When the wedge moves towards left, the sliding of the surfaces AC and AB will take place. At the same time load moves up and sliding of the load takes place along GD. Thus for the wedge and load shown in Fig.3.41 sliding takes place along surface AB, AC and DG. Hence there will be three normal reactions at AB, AC and DG.
The problems on wedges are generally the problems of equilibrium on inclined planes. Therefore, these problems are solved by equilibrium method or by applying Lami’s Theorem
Fig14
Equilibrium Method In this method, the equilibrium of the load (or the body placed on the wedge) and the equilibrium of the wedge are considered.
Equilibrium of Wedge
Consider the equilibrium of the wedge. The forces acting on the wedge are shown in Fig. 14. They are:
(i) The force P applied horizontally on face BC.
(ii) Reaction R1 on the face AC (The reaction R1 is the resultant of the normal reaction N1 on the rubbing face AB and force of friction on surface AC). The reaction R1 will be inclined at an angle θ1 (when θ1 is angle of friction) with the normal.
(iii) Reaction R2 on the face AB (The reaction R2 is the resultant of normal reaction N2 on the rubbing face AB and force of friction on surface AB). The reaction R2 will be inclined at an angle Φ2 with the normal.
When the force P is applied on the wedge, the surface CA will be moving towards left and hence force of friction on this surface will be acting towards right. Similarly, the force of friction on face AB will be acting from A to B. These forces are shown in Fig. 15.
Resolving the forces horizontally, we get
Fig 15 Equilibrium of wedge
Fig 16
Fig 17
Equilibrium of body placed on the Wedge
The forces acting on the body are shown in Fig.17. They are:
(i) The weight W on the body.
(ii)Reaction R3 on the face GD. (The reaction R3 is the resultant of the normal reaction N3 on the rubbing face GD and force of friction on surface GD).
(iii)Reaction R3 on the face GF (The reaction R2 is the resultant of the normal reaction N2 on the rubbing face GF and force of friction on surface GF).
These forces are shown in Fig. 17.
Resolving the forces R2, R3 and W horizontally, we get
If a ladder is placed against a rough horizontal floor and a vertical wall (smooth or roughly) then ladder is subjected to nonconcurrent force system.

Fig 18 A rough slope inclined at theta to the horizontal with a block of mass m resting on it. The block is in equilibrium so all of the forces acting on it sum to zero as shown by the vector triangle
Friction acts along the plane where the two surfaces are touching. For a block on a slope, as shown here, the friction force acts in the plane of the slope. In this case the block is stationary and so the frictional force acts to prevent it sliding down the slope, so the frictional force acts up the slope.
If the block is in equilibrium then the forces on it sum to zero. Resolving perpendicular to the slope gives N= mgcosθ. This means that the maximum frictional force isF=μmgcosθwhere μ is the coefficient of friction between the block and the slope.
A pulley is a device that can increase the magnitude of an effort force.
Fig 19
With no pulley  the effort force is similar to the load  in opposite direction.
S = F (1)
where
S = effort force (N, lb)
F = load (N, lb)
Single Pulley
Fixed Pulley
Fig 20
With a single fixed pulley the effort force is similar (or more due to efficiency loss) to the load.
S = F (2)
The advantage with the single fixed pulley is that the direction of force is changed  it is possible to pull down instead of lifting up.
Movable Pulley
Fig 21
With a single moveable pulley the effort force is half (or more due to efficiency loss) of the load.
S = 1/2 F (3)
Fig 22
With a combined moveable pulley as above  the effort force is half (or more due to efficiency loss) of the load.
S = 1/2 F (4)
Fig 23
With two pulleys and the ropes as above  the effort force is 1/3 (or more due to efficiency) of the load.
S = 1/3 F (5)
General Equation for Blocks and Tackles
The general effort force equation for a block or tackle to raise or pull a load can be expressed as
S = F / (μ n)
= (m g) / (μn) (6)
where
S = effort force (N, lb)
F = load (often weight) (N, lb)
m = mass (kg, slugs) (when lifting a mass)
g = constant of gravitation (9.81 m/s2, 32.17405 ft/s2) (when lifting a mass)
μ = mechanical efficiency of the system (equal to one for an ideal frictionless system, a fraction less than one for realworld systems with energy losses due to friction)
n = number of ropes between the sets of pulleys
References:
Andy Ruina and Rudra Pratap , Introduction to Statics and Dynamics, Oxford University Press.
Reddy Vijaykumar K. and K. Suresh Kumar, Singer's Engineering Mechanics.
F. P. Beer and E. R. Johnston, Mechanics for Engineers, Statics and Dynamics,
McGraw Hill. Irving H. Shames, Engineering Mechanics, Prentice Hall.