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A pull or push which changes or tends to change the state of rest or of uniform motion or direction of motion of an object is called Force.
Force is the interaction between the object and the source (providing the pull or push). It is a vector quantity. In this document vector quantities will be either denoted by bold letters like v or with bar on top .
Force: pull or push
There are basically two types of forces which are commonly encountered in mechanics.
Noncontact forces are the forces in which contact between two objects is not necessary.
Gravitational force between two bodies and electrostatic force between two charges are two examples of field forces.
Weight (W = mg) of a body comes in this category.
Contact force is a force that is applied by objects in contact with each other.
Consider two wooden blocks A and B being rubbed against each other.
In Figure shown below, A is being moved to the right while B is being moved leftward.
In order to see more clearly which forces act on A and which on B, a second diagram is drawn below showing a space between the blocks but they are still supposed to be in contact.
In Figure, the two normal reactions each of magnitude N are perpendicular to the surface of contact between the blocks and the two frictional forces each of magnitude fact along that surface, each in a direction opposing the motion of the block upon which it acts. We will read about Normal Reaction in detail later in the document.
Note: Forces on block B from the ground are not shown in the figure.
Newton has published three laws, which describe how forces affect motion of a body on which they act. These laws are fundamental in nature in the sense that the first law gives concept of force, inertia and the inertial frames; the second law defines force and the third law action and reaction as two aspects of mutual interaction between two bodies.
Every material body has tendency to preserve its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by external forces impressed on it.
The rate of change in momentum of a body is equal to, and occurs in the direction of the net applied force.
A body of mass m in translational motion with velocityif acted upon with a net external force F the second law suggests:
If mass of the body is constant, the above equation relates the acceleration a of the body with the net force F acting on it.
The first law provides concept of force and the second law provides the quantitative definition of force, therefore the second law is also valid only in inertial frames.
SI unit of force is newton. It is abbreviated as N. One newton equals to one kilogrammeter per second square.
1 N = 1 kgm/s^{2}
Dimensions of force are MLT^{–2}
Newton’s 3rd law of motion states that for every action, there is an equal and opposite reaction.
Force is always a twobody interaction. The first law describes qualitatively and the second law describes quantitatively what happens to a body if a force acts on it, but do not reveal anything about what happens to the other body participating in the interaction responsible for the force.
The third law accounts for this aspect of the force and states that every action on a body has equal and opposite reaction on the other body participating in the interaction.
A force on a body can only exists when there is another body to create it, therefore in every physical situation of concern there must be two or more bodies applying forces on each other. On the other hand the three laws of Newton, describe motion of a single body under action of several forces, therefore, to analyze a given problem, we have to consider each of the bodies separately one by one. This idea provides us with the concept of free body diagram.
A free body diagram is a pictorial representation in which the body under study is assumed free from rest of the system i.e. assumed separated from rest of the interacting bodies and is drawn in its actual shape and orientation and all the forces acting on the body are shown.
1. Identify the Contact Forces.
2. After identifying the contact forces, draw a dot to represent the object that we are interested in. Here, we are only interested in determining the forces acting on our object.
3. Draw a coordinate system and label positive directions.
4. Draw the contact forces on the dot with an arrow pointing away from the dot. The arrow lengths should be relatively proportional to each other. Label all forces.
5. Draw and label our longrange forces. This will usually be weight unless there is an electric charge or magnetism involved.
6. If there is acceleration in the system, then draw and label the acceleration vector.
Field forces include the gravitational force (weight) electrostatic forces and magnetic forces, which can easily be identified. At present, we consider only gravitational pull from the earth i.e. weight of the body.
The gravitational pull from the earth acts on every particle of the body hence it is a distributed force. The net gravitational pull of the Earth on a body may be considered as weight of the body.
At every point where a body under consideration is supposed to be separated from other bodies to draw its freebody diagram, there may be a contact force. Most common contact forces, which we usually encounter, are tension force of a string, normal reaction on a surface in contact, friction, spring force etc.
A pulley is a device consisting of a wheel, which can rotate freely on its axel. A single pulley changes direction of tension force. At present for simplicity, we discuss only ideal pulley, which is massless i.e. has negligible mass and rotates on its axel without any friction. An ideal pulley offers no resistance to its rotation, therefore tension force in the string on both sides of it are equal in magnitude. Such a pulley is known as ideal pulley.
Two bodies in contact, when press each other, must apply equal and opposite forces on each other. These forces constitute a third law actionreaction pair. If surfaces of the bodies in contact are frictionless, this force acts along normal to the surface at the point of contact. Therefore, it is known as normal reaction.
Consider a block of weight W placed on a frictionless floor. Because of its weight it presses the floor at every point in contact and the floor also applies equal and opposite reaction forces on every point of contact. We show all of them by a single resultant N obtained by their vector addition.
Normal Reaction
To apply Newton's laws of motion (NLM) on the block, its weight W and normal reaction N applied by the floor on the block must be considered as shown in the following figure. It is the FBD of the block.
Consider a spherical ball of weight W placed on a floor. The normal reaction from the floor on the ball and from the ball on the floor makes a third law actionreaction pair. These forces are shown in the left figure.
To apply Newton's laws of motion (NLM) on the ball, its weight W and normal reaction N applied by the floor on the ball must be considered as shown in the above right figure. It is the FBD of the ball.
When two surfaces make contact, the normal reaction acts along the common normal and when a surface and a sharp corner make a contact the normal reaction acts along the normal to the surface. Consider a block placed in a rectangular trough as shown in the figure.
To apply Newton's laws of motion (NLM) on the block, its free body diagram (FBD) is shown in the above right figure.
When no force acts on a spring, it is in relaxed condition i.e. neither compressed nor elongated. Consider a spring attached to a fixed support at one of its end and the other end is free. If we neglect gravity, it remains in relaxed state. When it is pushed by a force F, it is compressed and displacement x of its free end is called compression. When the spring is pulled by a force F, it is elongated and displacement x of its free end is called elongation. Various forces developed in these situations are shown in the following figure.
Various cases of spring force
The force applied by the spring on the wall and the force applied by the wall on the spring make a third law actionreaction pair. Similarly, force by hand on the spring and the force by spring on the hand make another third law actionreaction pair.
Forces which have zero resultant and zero turning effect will not cause any change in the motion of the object to which they are applied. Such forces (and the object) are said to be in equilibrium. For understanding the equilibrium of an object under two or more concurrent or coplanar forces let us first discuss the resolution of force and moment of a force about some point.
When a force is replaced by an equivalent set of components, it is said to be resolved. One of the most useful ways in which to resolve a force is to choose only two components (although a force may be resolved in three or more components also) which are at right angles also. The magnitude of these components can be very easily found using trigonometry.
In Fig. 8.17, F_{1} = F cos θ = component of F along AC.
F_{2} = F sin θ = component of F perpendicular to AC AB.
Finding such components is referred to as resolving a force in a pair of perpendicular directions.
Note: that the component of a force in a direction perpendicular to itself is zero. For example, if a force of 10 N is applied on an object in horizontal direction then its component along vertical is zero. Similarly, the component of a force in a direction parallel to the force is equal to the magnitude of the force. For example component of the above force in the direction of force (horizontal) will be 10 N.
In the opposite direction the component is −10 N.
Example 8.4 Resolve a weight of 10 N in two directions which are parallel and
perpendicular to a slope inclined at 30° to the horizontal.
Solution: Component perpendicular to the plane
and component parallel to the plane
Example 8.5 Resolve horizontally and vertically a force F N = 8 which makes
an angle of 45° with the horizontal.
Solution Horizontal component of F is
= 4√2 N
and vertical component of F is F_{V} = ° sin 45
Example 8.6 A body is supported on a rough plane inclined at 30° to the horizontal by a string attached to the body and held at an angle of 30° to the plane. Draw a diagram showing the forces acting on the body and resolve each of these forces
(a) horizontally and vertically,
(b) parallel and perpendicular to the plane.
Solution. The forces are:
(i) The tension in the string T
(ii) The normal reaction with the plane N
(iii) The weight of the body w and the friction f
(a) Resolving horizontally and vertically
Resolving horizontally and vertically in the senses OX and OY as shown, the components are
(b) Resolving parallel and perpendicular to the plane
Resolving parallel and perpendicular to the plane in the senses OX ′ and OY ′ as shown, the components are :
The general name given to any turning effect is torque. The magnitude of torque, also known as the moment of a force F is calculated by multiplying together the magnitude of the force and its perpendicular distance from the axis of rotation. This is denoted by C or π (tau). i.e
The angular direction of torque is the sense of the rotation it would cause.
Consider a lamina that is free to rotate in its own plane about an axis perpendicular to the lamina and passing through a point A on the lamina. In the diagram, the moment about the axis of rotation of the force F_{1} is F_{1} r_{1} anticlockwise and the moment of the force F_{2}
is F_{2}r_{2} clockwise. A convenient way to differentiate between clockwise and anticlockwise torques is to allocate a positive sign to one sense (usually, but not invariably, this is anticlockwise) and a negative sign to the other. With this convention, the moments of F_{1}
and F_{2} are + F_{1}r_{1} and −F_{2} r_{2} (when using a sign convention in any problem it is advisable to specify the chosen positive sense).
If the line of action of a force passes through the axis of rotation, its perpendicular distance from the axis is zero. Therefore, its moment about that axis is also zero.
Note: Later in the chapter of rotation we will see that torque is a vector quantity.
Example 8.7 ABCD is a square of side 2 m and O is its centre. Forces act along the sides as shown in the diagram.
Calculate the moment of each force about
(a) an axis through A and perpendicular to the plane of square.
(b) an axis through O and perpendicular to the plane of square.
Solution. Taking anticlockwise moments as positive we have:
Example 8.8 Forces act as indicated on a rod AB which is pivoted at A. Find
the anticlockwise moment of each force about the pivot.
Solution.
When an object is in equilibrium under the action of a set of two or more coplanar forces, each of three factors which comprise the possible movement of the object must be zero, i.e. the object has
(i) no linear movement along any two mutually perpendicular directions OX and OY.
(ii) no rotation about any axis.
The set of forces must, therefore, be such that
(a) the algebraic sum of the components parallel to OX is zero or ∑F_{x} = 0
(b) the algebraic sum of the components parallel to OY is zero or ΣF_{y} = 0
(c) the resultant moment about any specified axis is zero or Στ_{any axis} = 0
Thus, for the equilibrium of a set of two or more coplanar forces
Using the above three conditions, we get only three set of equations. So, in a problem number of unknowns should not be more than three.
Example 8.9 A rod AB rests with the end A on rough horizontal ground and the end B against a smooth vertical wall.
The rod is uniform and of weight w. If the rod is in equilibrium
in the position shown in figure. Find
(a) frictional force at A
(b) normal reaction at A
(c) normal reaction at B.
Solution Let length of the rod be 2l. Using the three conditions of
equilibrium. Anticlockwise moment is taken as positive.
(i) ΣF_{X} = 0 ⇒ ∴ N_{B} f_{A} − = 0
or N_{B} =f_{A} = ...(i)
(ii) ΣF_{Y} = 0 ⇒ ∴ N_{A} − w = 0
or N_{A }= w ... (ii)
(iii) Στ_{O} = 0
∴ N_{A}(2l cos 30° )  N_{B} ( 2Isin 30° )  w (l cos 30° ) = 0
or ...(iii)
Solving these three equations, we get
If an object is in equilibrium under two or more concurrent coplanar forces the algebraic sum of the components of forces in any two mutually perpendicular directions OX and OY should be zero, i.e. the set of forces must be such that
(i) the algebraic sum of the components parallel to OX is zero, i.e. ΣF_{x} = 0.
(ii) the algebraic sum of the components parallel to OY is zero, i.e. ΣF_{y} = 0.
Thus, for the equilibrium of two or more concurrent coplanar forces
The third condition of zero moment about any specified axis is automatically satisfied if the moment is taken about the point of intersection of the forces. So, here we get only two equations. Thus, number of unknown in any problem should not be more than two.
Example 8.10 An object is in equilibrium under four concurrent forces in the
directions shown in figure. Find the magnitudes of F_{1} and F_{2}.
Solution: The object is in equilibrium. Hence,
(i) ΣFx = 0
∴ 8 + 4cos 60° F_{2} cos 30°
or
or
(ii) ΣF_{y} = 0
F_{1}+ 4sin 60°− F_{2} sin 30° = 0
or
or
or
If an object O is in equilibrium under three concurrent forces F_{1}, F_{2} and F_{3} as shown in figure. Then,
This property of three concurrent forces in equilibrium is known as Lami’s theorem and is very useful method of solving problems related to three concurrent forces in equilibrium.
Example 8.11 One end of a string 0.5 m long is fixed to a point A and the other end is fastened to a small object of weight 8 N. The object is pulled aside by a horizontal force F, until it is 0.3 m from the vertical through A. Find the magnitudes of the tension T in the string and the force F.
Solution AC = 0.5 m, BC = 0.3 m
∴ AB = 0.4 m
and if ∠BAC= θ.
Then
Here, the object is in equilibrium under three concurrent forces.
So, we can apply Lami’s theorem.
or
∴
Example 8.12 The rod shown in figure has a mass of 2 kg and length 4 m. In equilibrium, find the hinge force (or its two components) acting on the rod and tension in the string. Take g = 10m/s^{2}.
Solution
In the figure, only those forces which are acting on the rod has been shown. Here H and V are horizontal and vertical components of the hinge force.
Σ F_{x }= 0 ⇒ H T − = 0.8 0 ...(i)
Σ F_{y} = 0 ⇒ V + 0.6T − 20 = 0 ...(ii)
Στ_{O} = 0
⇒ Clockwise torque of 20 N = anticlockwise torque of 0.6 T.
All other forces ( , H V and 0.8 T pass through O, hence their torques are zero).
∴ 20 × 2 = 0.6T X 4(iii)
Solving these three equations, we get
T = 16.67 N,
H = 13.33 N
and V = 10 N
Hinge force (F)
Ans.
∴ Ans.
When Subjected to a Horizontal Pull
The distribution of forces on the body are shown. As there is no motion along vertical direction, hence, R = mg
For horizontal motion F = ma or a = F/m
When Subjected to a Pull acting at an Angle (θ) to the Horizontal
Now F has to be resolved into two components, F cosθ along the horizontal and F sinθ along the vertical direction.
For no motion along the vertical direction.
we have R F sinθ = mg
or R = mg F sinθ
For horizontal motion
F cosθ = ma, a =
When the Block is Subjected to a Push Acting at an Angle θ to the Horizontal (Downward)
The force equation in this case
R = mg F sinθ
For horizontal motion
F cosθ = ma, a =
Two Body System
Let a force F be applied on mass m_{1}
Free body diagrams:
(vertical force do not cause motion, hence they have not been shown in diagram)
⇒ a = and f =
(i) Here f is known as force of contact.
(ii) Acceleration of system can be found simply by
a =
: If force F be applied on m_{2}, the acceleration will remain the same, but the force of contact will be different
i.e., f' =
Ques 6: Find the contact force between the 3 kg and 2kg block as shown in figure.
Ans: Considering both blocks as a system to find the common acceleration
F_{net} = F_{1} F_{2} = 100 25 = 75 N
common acceleration
To find the contact force
between A & B we draw
F.B.D of 2 kg block
from (åF_{net})_{x} = ma_{x}
⇒ N 25 = (2) (15)
⇒ N = 55 N
Three Body System
Free body diagrams:
⇒ a =
and f_{1} =
f_{2} =
f_{1} = contact force between masses m_{1} and m_{2}
f_{2} = contact force between masses m_{2} and m_{3}
Remember: Contact forces will be different if force F will be applied on mass C
Ques 7: Find the contact force between the block and acceleration of the blocks as shown in figure.
Ans: Considering all the three block as a system to find the common acceleration
F_{net} = 50 30 = 20 N
To find the contact force
between B & C we draw F.B.D.
of 3 kg block.
⇒ N_{1} 30 = 3(2) ⇒ N_{1} = 36 N
To find contact force between A & B we draw
F.B.D. of 5 kg block
⇒ N_{2} N_{1} = 5a
N_{2} = 5 × 2 36 ⇒ N_{2} = 46 N
For Two Bodies
F is the pull on body A of mass m_{1}. The pull of A on B is exercised as tension through the string connecting A and B. The value of tension throughout the string is T only.
Free body diagrams :
⇒ a =
Case (ii) : For Three bodies:
Free body diagrams:
⇒ a =
Ques 8: A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward is 2 m/s^{2} by an external force F_{0}.
(a) What is F_{0}?
(b) What is the force on rope?
(c) What is the tension at middle point of the rope?
(g = 10 m/s^{2})
Ans: For calculating the value of F_{0}, consider two blocks with the rope as a system.
F.B.D. of whole system
(a)
F_{0} 100 = 10 × 2
F = 120 N ...(1)
(b) According to Newton's second law, net force on rope.
F = ma = (2) (2) = 4 N ...(2)
(c) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown.
T 4g = 4.(2) = 48 N
Motion of a body on a smooth inclined plane:
Natural acceleration down the plane = g sinθ
Driving force for acceleration a up the plane, F=m(a gsinθ)
and for an acceleration a down the plane, F=m(a gsinθ)
Ques 9: Find out the contact force between the 2kg & 4kg block as shown in figure.
Ans: On an incline plane acceleration of the block is independent of mass. So both the blocks will move with the same acceleration (gsin 37º) so the contact force between them is zero.
Ques 10: Find out the contact force between 2kg & 3kg block placed on the incline plane as shown in figure.
Ans: Considering both the block as a 5 kg system because both will move the same acceleration.
Now show forces on the 5 kg block:
Acceleration of 5 kg block is down the incline.
So choose one axis down the incline and other
perpendicular to it
From Newton's second Law
N = 5g cos 37º ...(i)
5gsin 37º 20 = 5a ..(ii)
30 20 = 5a
a = 2m/s^{2} (down the incline)
For contact force (N_{1}) between 2 kg & 3 kg block
we draw F.B.D. of 3 kg block
From
F_{net} = ma
⇒ 3 g sin 37º N_{1} = 3 × 2
18 N_{1} = 6
N_{1}= 12 N
Ques 11: One end of string which passes through pulley and connected to 10 kg mass at other end is pulled by 100 N force. Find out the acceleration of 10 kg mass. (g = 9.8 m/s^{2})
Ans: Since string is pulled by 100N force.
So tension in the string is 100 N.
F.B.D. of 10 kg block
100 10 g = 10 a
100 10 × 9.8 = 10 a
a = 0.2 m/s^{2}
Ques 12: In the figure shown, find out acceleration of each block.
Ans: Now F.B.D. of each block and apply Newton's second law on each F.B.D
(1) ⇒ 10g 2 T = 10a_{1} ...(i)
(2) ⇒ T 2g = 2a_{2 } ...(2)
(3) ⇒ T 4g = 4a_{3} ...(3)
from constrain relation 2a_{1} = a_{2} + a_{3} ...(4)
Solving equations (1), (2), (3) and (4) we get
T =
a_{1} = 70/23 m/s^{2 }(downward), a_{2} = 170/23 m/s^{2 }(upward), a_{3} = 30/23 m/s^{2 }(downward)
Ques 13: Find the acceleration of each block in the figure shown below; in terms of their masses m_{1}, m_{2} and g. Neglect any friction.
Ans: Let T be the tension in the string that is assumed to be massless.
For mass m_{1}, the FBD shows that
N_{1} = m_{1}g
Where N_{1} is the force applied upward by plane on the mass m_{1}.
If acceleration of m_{1} along horizontal is a_{1}. then
T = m_{1}a_{1} ...(i)
For mass m_{2}, the FBD shows that
m_{2}g 2T = m_{2}a_{2} ...(ii)
Where a_{2} is vertical acceleration of mass m_{2}.
Note that upward tension on m_{2} is 2T applied by both sides of the string.
from constrain relation
a_{2} =
Thus, the acceleration of m_{1} its twice that of m_{2}.
with this input, solving (i) and (ii) we find
a_{1} =
a_{2} =
Ques 14: Two blocks A and B each having a mass of 20 kg, rest on frictionless surfaces as shown in the figure below. Assuming the pulleys to be light and frictionless, compute:
(a) the time required for block A, to move down by 2m on the plane, starting from rest,
(b) tension in the string, connecting the blocks.
Ans:
Step 1: Draw the FBDs for both the blocks. If tension in the string is T, then we have
and
Note that m_{A}g, should better be resolved along and perpendicular to the plane, as the block A is moving along the plane.
Step 2: From FBDs, we write the force equations `
for block A where
N_{A} = m_{A} g cosθ = 20 × 10 × 0.8 = 160 N
and m_{A}g sinθ T = m_{A} a ... (i)
Where `a' is acceleration of masses of blocks A and B. Similarly, force equations for block B are
N_{B} = m_{B}g = 20 × 10 = 200 N
and T = m_{B}a ...(ii)
From (i) and (ii), we obtain
a = = = 3 ms^{2}
T = m_{B}a = 20 × 3 = 60 N
Step 3: With constant acceleration a = 3 ms^{2}, the block A moves down the inclined plane a distance S = 2 m in time t given by
S = or t =
Ques 15: Two blocks m_{1} and m_{2} are placed on a smooth inclined plane as shown in figure. If they are released from rest.
Find:
(i) acceleration of mass m_{1} and m_{2 }
(ii) tension in the string
(iii) net force on pulley exerted by string
Ans: F.B.D of m_{1} :
m_{1}g sinθ T = m_{1}a
T = ...(i)
F.B.D. of m_{2} :
T m_{2}g sinθ = m_{2}a
T 1. = 1.a ...(ii)
Adding eq. (i) and (ii) we get a = 0
Putting this value in eq. (i) we get
T = ,
F.B.D. of pulley
F_{R} =
F_{R} =
Ques 16: A block of mass `m' is kept on the ground as shown in figure.
(i) Draw F.B.D. of block
(ii) Are forces acting on block action  reaction pair
(iii) If answer is no, draw action reaction pair.
Ans: (i) F.B.D. of block
(ii) `N' and Mg are not action  reaction pair. Since pair act on different bodies, and they are of same nature.
(iii) Pair of `mg' of block acts on earth in opposite direction.
and pair of `N' acts on surface as shown in figure.
Climbing on the Rope
Now three condition arises.
if T > mg ⇒ man accelerates in upward direction
T < mg ⇒ man accelerates in downward direction
T = mg ⇒ man's acceleration is zero
Either climbing or descending on the rope man exerts force downward
Ques 17 If the breaking strength of string is 600N then find out the maximum acceleration of the man with which he can climb up the road
Ans: Maximum force that can be exerted on the man by the rope is 600 N.
F.B.D of man
⇒ 600 50 g = 50 a
a_{max} = 2 m/s^{2}
⇒ 600 50 g = 50 a
a_{max} = 2 m/s^{2}
Ques 18: A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform.
(i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.
(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s?
(iii) What force should apply now to maintain the constant speed of 1 m/s?
Ans: The free body diagram of the painter and the platform as a system can be drawn as shown in the figure. Note that the tension in the string is equal to the force by which he pulled the rope.
(i) Applying Newton's Second Law
2T (M m)g = (M m)a
or a =
Here M = 60 kg; m = 15 kg ; T = 400 N
g = 10 m/s^{2}
a = = 0.67 m/s^{2}
(ii) To attain a speed of 1 m/s in one second the acceleration a must be 1 m/s^{2}
Thus, the applied force is
F = (M m) (g a) = (60 15) (10 1) = 412.5 N
(iii) When the painter and the platform move (upward) together with a constant speed, it is in a state of dynamic equilibrium
Thus, 2F (M m) g = 0
or F = = 375 N
Every spring resists any attempt to change its length; when it is compressed or extended, it exerts force at its ends. The force exerted by a spring is given by F = kx, where x is the change in length and k is the stiffness constant or spring constant (unit Nm^{1})
When spring is in its natural length, spring force is zero.
(a) If block `A' and `B' both are displaced by 0.5 m in same direction.
(b) If block `A' and `B' both are displaced by 0.5 m in opposite d.
Before studying the concept of pseudo force let us first discuss frame of reference.Frame of reference is the way of observation the things.
A nonaccelerating frame of reference is called an inertial frame of reference. A frame of reference moving with a constant velocity is an inertial frame of reference.
An accelerating frame of reference is called a noninertial frame of reference.
Note: (i) A rotating frame of reference is a noninertial frame of reference, because it is also an accelerating one.
(ii) Earth is rotating about its axis of rotation and it is revolving around the centre of sun also. So, it is nonintertial frame of reference. But for most of the cases, we consider its as an inertial frame of reference. Now let us come to the pseudo force. Instead of ground (or inertial frame of reference) when we start watching the objects from a noninertial (accelerating) frame of reference its motion conditions are felt differently.
For example: Suppose a child is standing inside an accelerating lift. From ground frame of reference this child appears to be accelerating but from lift (noninertial) frame of reference child appears to be at rest. To justify this changed condition of motion, from equations point of view we have to apply a pseudo force. This pseudo force is given by
Here, ‘m’ is the mass of that body/object which is being observed from noninertial frame of reference and a is the acceleration of frame of reference. Negative sign implies that direction of pseudo force F_{p} is opposite to a. Hence whenever you make free body diagram of a body from a noninertial frame, apply all real forces (actually acting) on the body plus one pseudo force. Magnitude of this pseudo force is ‘ma’ and the direction is opposite to a.
Example: Suppose a block A of mass m is placed on a lift ascending with an acceleration a_{0}. Let N be the normal reaction between the block and the floor of the lift. Free body diagram of A in ground frame of reference (inertial) is shown in Fig. 8.100.
∴
But if we draw the free body diagram of A with respect to the elevator (a noninertial frame of reference) without applying the pseudo force, as shown in Fig. 8.101, we get
or N mg ′ = ...(ii)
Since, N ≠ N, either of the equations is wrong. If we apply a pseudo force in noninertial frame of reference, N ′ becomes equal to N as shown in Fig. 8.102.
Acceleration of block with respect to elevator is zero.
∴ N' mg  ma_{0 }= 0
or N' m= (g + a_{0}) ...(iii)
∴ N' = N
Example 8.24 All surfaces are smooth in following figure. Find F, such that block remains stationary with respect to wedge.
Solution Acceleration of (block + wedge)
Let us solve the problem by both the methods. Such problems can be solved with or without using the concept of pseudo force.
From Inertial Frame of Reference
FBD of block w.r.t. ground (Apply real forces):
With respect to ground block is moving with an acceleration a.
∴ ΣF_{y} = 0
⇒ N mg cos θ = ...(i)
and ΣF_{x} = ma
⇒ N sin θ = ma...(ii)
From Eqs. (i) and (ii), we get
a = g tan θ
∴ F= (M+ m)a
= (M+ m) g tan θ
From Noninertial Frame of Reference (Wedge)
FBD of block w.r.t. wedge (real forces + pseudo force) w.r.t. wedge, block is stationary
∴ ΣF_{y} = 0 ⇒ N mg cos θ = ...(iii)
ΣF_{x} = 0 ⇒ N ma sin θ = ...(iv)
From Eqs. (iii) and (iv), we will get the same result
i.e. F= (M+ m) g tan θ
Example 8.25 A bob of mass m is suspended from the ceiling of a train moving with an acceleration a as shown in figure. Find the angle θ in equilibrium position.
Solution This problem can also be solved by both the methods.
Inertial Frame of Reference (Ground)
FBD of bob w.r.t. ground (only real forces)
With respect to ground, bob is also moving with an acceleration a.
∴ ΣF_{x} = 0 ⇒ T ma sin θ = ...(i)
and ΣF_{y} = 0 ⇒ T mg cos θ = ...(ii)
From Eqs. (i) and (ii), we get
Noninertial Frame of Reference (Train)
FBD of bob w.r.t. train (real forces + pseudo force):
with respect to train, bob is in equilibrium
∴ ΣF_{x }= 0
⇒ T sin θ = ...(iii)
∴ ΣF_{y} = 0
⇒ T cos θ = ...(iv)
From Eqs. (iii) and (iv), we get the same result, i.e.
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