Topics

Mass is that property of a substance which specifies how much resistance an object exhibits to changes in its velocity i.e. it gives an idea of inertia of that body

Suppose a ball is moving in a straight line and a train is moving in a straight line; both with same constant speed. Which one is more difficult to stop? Of course the train.

Mass is an inherent property of an object and of the method used to measure it. The mass of a closed system of bodies is independent of the processes going on in the system, no matter what kind these processes are. Also, mass is a scalar quantity and thus obeys the rules of ordinary arithmetic.

Mass should not be confused with weight. Mass and weight are two different quantities. Weight is the force with which Earth attracts the object and is dependent on mass.

Newton's Second law

Newton's first law explains what happens to an object when no net force acts on it. It either remains at rest or moves in straight line with constant speed. Newton's second law answers the question, what happens to an object that has a nonzero resultant force acting on it.

When viewed from an inertial reference frame, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

Force of hand accelerates the block

The same force accelerates 2 blocks \(1/2\) as much acceleration

acceleration \(\propto \frac{1}{\text{~}\text{mass}\text{~}}\)
Thus, we can relate mass, acceleration, and force through the following mathematical statement of Newton's second law, by choosing proper units so that constant of proportionality is 1 :

\[\begin{array}{r} \sum_{}^{}\ \overrightarrow{F} = m\overrightarrow{a}\#(i) \end{array}\]

we have indicated that the acceleration is due to the net force acting on an object. The net force on an
object is the vector sum of all forces acting on the object. In solving a problem using Newton's second law, it is important to determine the correct net force on an object.

What happens when several forces act simultaneously on an object?
In this case, the object accelerates only if the net force acting on it is not equal to zero.

\[\begin{array}{r} {\overrightarrow{F}}_{\text{net}\text{~}} = {\overrightarrow{F}}_{1} + {\overrightarrow{F}}_{2} + {\overrightarrow{F}}_{3} + \ldots\ldots..:\text{~}\text{vector sum of all forces.}\text{~}\#(ii) \end{array}\]

There may be many forces acting on an object, but there is only one acceleration.
Note that equation 2 is vector expression and hence is equivalent to three component equations

\[\sum_{}^{}\ F_{x} = {ma}_{x}\ \sum_{}^{}\ {\text{ }F}_{y} = {ma}_{y}\ \sum_{}^{}\ {\text{ }F}_{z} = {ma}_{z}\]

When net force is zero, then the object is said to be in equilibrium. And if net force is zero, acceleration will also be zero, therefore velocity will be constant. When the velocity of an object is constant (including when object is permanently at rest), the object is said to be in equilibrium.

Force is the cause of change in motion :

Force does not cause motion. We can have motion in the absence of force, as described in Newton's first law. Force is the cause of change in motion as measured by acceleration.

"ma" is not a force :

Equation (i) does not say that the product ma is a force. All forces on object are added vectorialy as in equation (ii) to get the net force on the left side of the equation. This net force is then equated to the product of the mass of object and the acceleration that results from the net force.

Do not include an "ma force" in your analysis of the forces on an object.

Definition of the newton

The SI unit of force is the newton, which is defined as the force that, when acting on an object of mass 1 kg , produces an acceleration of \(1\text{ }m/s^{2}\). From the definition and Newton's second law, we see that the newton can be expressed in terms of the following fundamental units of mass, length, and time
\[1\text{ }N \equiv 1\text{ }kg.m/s^{2}\]

Newton's Third law

Every action has equal and opposite reaction, and the action reaction act on the different bodies, simultaneously These action reaction pair must be of same type, viz. friction for friction, gravitation for gravitation etc.

Caution

(i) Action and reaction forces act on different objects. Two forces acting on the same object, even if they are equal in magnitude and opposite in direction, cannot be an action-reaction pair.
(ii) In an action reaction pair, both forces act simultaneously i.e. we can not say one as action and other reaction.

Practice Exercise

Q. 1 An object experiences no acceleration. Which of the following cannot be true for the object?
(A) A single force acts on the object.
(B) No force acts on the object.
(C) Forces act on the object, but the forces cancel.
Q. 2 An object experiences net force and exhibits an acceleration in response. Which of the following statements is always true?
(A) the object moves in the direction of the force.
(B) The acceleration is in the same direction as the velocity.
(C) The acceleration is in the same direction as the net force.
(D) The velocity of the object increases.
Q. 3 If a fly collides with the windshield of a fast-moving bus, which object experiences an impact force with a larger magnitude?
(A) the fly
(B) the bus
(C) the same force is experienced by both.
Q. 4 If a fly collides with the windshield of a fast-moving bus, which object experiences the greater acceleration:
(A) The fly
(B) the bus
(C) the same acceleration is experienced by both.

In this diagram the object of interest is isolated from its surroundings and the interactions between the object and the surroundings are represented in terms of forces.

Remember not to show " \(M\overrightarrow{a}\) " as force in FBD. \(\overrightarrow{a}\) is the effect of net force acting on an object.

System

The system is what you choose to analyse motion.

1. According to system chosen, forces can be divided into two categories :
   (i) Internal force
   (ii) External force

Whether a force is external or internal will depend on the system chosen. For example, earth is pulling a mass by applying a force mg on it. The mass is also applying a force mg on earth.
If we chose earth-mass as system, then mg becomes an internal force as it is within the system. But if choose only mass \(m\) as system; then \(mg\) acting on the mass is an external force.

(system)

We do not show internal force in FBD while solving a problem.
2. According to the origin of force, we can classify it into two types :
(i) Field forces
(ii) Contact forces

Field forces

The force experienced by an object without physical contact is known as field force.
e.g. Gravitational force, Electromagnetic force.

Gravitational force :

The force by which two bodies attract each other by virtue of their masses.
If two particles haivng masses \(m_{1}\& m_{2}\) are seperated by a distance \(r\), then the magnitude of force is given by \(F = \frac{{Gm}_{1}{\text{ }m}_{2}}{r^{2}}\).

This is called Newton's law of Gravitation
The force acts along the line joining the particles, this force is always attractive in nature.
For earth mass system, earth applies a force mg on mass m , which always acts towards the center of earth downwards. mg is called weight of the body.

Electromagnetic force :

If two particles having charges \(q_{1}\) and \(q_{2}\) are separated by a distance r , then the magnitude of force is given by : \(F = \frac{Kq_{1}q_{2}}{r^{2}}\). This is called Coulomb's law and it acts along the line joining the particles. For like charges, the force is repulsive, and for unlike charges, the force is attractive.

Contact Force

The force experienced by an object by physical contact is known as contact force.
e.g. Frictional force (which will be dealt in detail later), Normal force, tension.

Normal force :

A contact force perpendicular (normal means perpendicular) to the contact surface that prevents two objects from passing through one another is called the normal contact force.
In the situation shown, LCD is kept on the table. Weight mg acts on it downwards still it is at rest. i.e. some force acts in the upward direction (Normal to the table) which balances the weight. This force is the normal force by the table on the LCD.
\(F_{mT} \Rightarrow F\) on mass by table
\(F_{me} \Rightarrow F\) on mass by Earth

(b)

Here \(N = mg\) (not an action reaction pair)

N Does not always equal mg

In the situation shown in figure and in many others, we find that the normal force has the same magnitude as the gravitational force. However, this is not always true. If an object is on an incline, if there are applied forces with vertical components, or if there is a vertical acceleration of the system, then \(N \neq mg\). Always apply Newton's second law to find the relationship between N and mg .
Note : Normal force is always a pushing force.
Normal force acts along the common normal
e.g. (i)

e.g. (ii)

\(N_{1}\) : Normal force between m & M

\(N_{2}\) : Normal force between M & floor

Normal force acts perpendicular to the surface in contact.

If one surface is not well defined, then normal force acts perpendicular to the surface of second object.
Here we can not specity the perpenducular to the rod at the point of contact. Thus normal acts perpenducular to the floor and the wall respectively.
e.g.

FBD of A

\[\begin{matrix} 2\text{ }Ncos30^{\circ} = mg \\ \text{ }N = \frac{mg}{\sqrt{3}} \end{matrix}\]

Considering all the three together as system,

\[\begin{matrix} & 2{\text{ }N}^{'} = 3mg \\ & {\text{ }N}^{'} = \frac{3mg}{2} \end{matrix}\]

Practice Exercise

Q. 1 Find normal forces at all the contact points.
(i)

(ii)

Q. 2 Find normal contact force between sphere and inclind plane.

Answers

Q. 1 (i) Normal contact force between 5 kg and 10 kg is 95 N . while 10 kg and 35 kg is 85 N .
(ii) \(\frac{100}{\sqrt{3}}\text{ }N\)
Q. 23 mg

• It is self adjusting pulling force, which is electromagnetic in nature.
  -We assume string is massless, Unless specified.

• Strings will be tight only when the ends are pulled apart.

• Tension is a pulling force, acts away from the object along the string

• Tension throughout the massless string is same usually.

• A pulley can change the direction of the force exerted by a cord.

(a)

When \(m_{l} = m_{2^{'}},a = 0\) and \(T = mg = m_{1}g = m_{2}g\)
If \(m_{2} \gg m_{p},a \approx g\) (a freely falling body) and \(T \approx 2m_{p}g\).

Basic steps for applying Newton's Laws

The following procedure is recommended when dealing with problems involving Newton's laws:

• Draw a simple, neat diagram of the system to help conceptualize the problem.

• If any acceleration component is zero, the particle is in equilibrium in this direction and \(\Sigma F = 0\) in this direction. If not the particle is undergoing an acceleration, the problem is one of non-equilibrium in this direction and \(\Sigma F = ma\).

• Isolate the object whose motion is being analyzed. Draw a free-body diagram for this object. For systems containing more than one object, draw separate free-body diagrams for each object.

Do not include in the free-body diagram forces exerted by the object on its surroundings.

• Establish convenient coordinate axes for each object and find the components of the forces along these axes. Apply Newton's second law, \(\Sigma F = ma\), in component form. Check your dimensions to make sure that all terms have units of force.

• Solve the component equations for the unknowns. Remember that you must have as many independent equations as you have unknowns to obtain a complete solution.

• Make sure, your results are consistent with the free-body diagram. Also check the predictions of your solutions for extreme value of the variables. By doing so you can often detect errors in your results.

Practice Exercise

Q. 1 Find the accelerations of blocks \(A,B\) & C and tentions in the strings. All the blocks are of equal mass 10 kg and \(F = 60\text{ }N\)

Q. 2 In figure \(m_{1} = 2\text{ }kg,m_{2} = 5\text{ }kg\) and \(F = 21\text{ }N\). Find the acceleration of blocks.

Q. 3 Find the acceleration of the 500 g block in figure.

Q. 4 Find the accelerations of blocks \(A,B\) & C and tensions in the strings. All the blocks are of equal mass 10 kg and \(F = 300\text{ }N\)

Q.5 A body of mass 12.5 kg is suspended with the help of light inextensible strings as shwon in figure. Find tension in three strings. Strings are light \(\left\lbrack g = 10{\text{ }ms}^{- 2} \right\rbrack\)

Q. 6 Which of the following is the reaction force to the gravitational force acting on the body as you sit in your desk chair?
(a) The normal force exerted by the chair
(b) The force you exert downward on the seat of the chair
(c) Neither of these forces.

Answers

We know that the more force we apply to a spring, the more it stretches. For a spring that obeys Hooke's low, the extension of the spring is proportional to the applied force. If we stretch spring by a distance x from its equilibrium position, it applies a restoring force F , towards its equilibrium position, which is proportional to \(x\), given by

\[F = kx\]

Here k is a proportionality constant, known as spring constant or stiffness. A spring has tendecy of restoring its natural length state, thus whether we stretch it or compress, it always opposes the external force in the direction towards its equilibrium position.

One more point is to be noted is that, a spring applies restoring force equally at both of its ends, doesn't matter whether an end is fixed or not.
If we look at FBD of spring we will note that equal force on spring must act from both ends.

\[F - F^{'} = m_{s} \times a\ \left( \text{~}\text{spring is mass less,}\text{~}m_{s} \rightarrow 0 \right)\]

therefore \(\ F = F^{'}\)

In figure an end of the spring is fixed to wall and other is pulled by applying a force. As the restoring force is directly proportional to the deformation in it, for stretching it by x , we apply a force F on it and for stretching it to double the length \((2x)\), we have to apply a force double of the previous value i.e. 2 F .

Let's say Ram & Shyam are pulling a spring from two ends as shown in figure Ram moves \(x_{2}\) and Shyam moves \(x_{1}\).

The force acting on Ram and Shyam is \(k\left( x_{1} + x_{2} \right)\), not \({kx}_{2}\) on Ram and \({kx}_{1}\) on Shyam. Force due to spring is kx where x is defined as \(\left| l - l_{0} \right|\), where \(l\) is present length and \(l_{0}\) is natural length,

(In all the discussions we will ignore mass and and damping of spring)
When two or more springs are connected in some manner then the combination can be replaced by a single equivalent spring such that it produces same elongation for same applied force.

Parallel combination

When springs are connected in parallel, then we can replace them by single spring of spring constant \(k_{eq}\) where \(k_{eq} = k_{1} + k_{2}\). This situation is shown in figure. Here we present the proof.
If the force F pulls the mass m by y , the stretch in each spring will be same.

\[y_{1} = y_{2} = y\]

Now for an equivalent spring \(F = k_{eq} \times y\) and as spring constants are not equal so \(F_{1} \neq F_{2}\).
For equivalence,

\[F = F_{1} + F_{2} \Rightarrow K_{eq}y = k_{1}y + k_{2}y\]

this reduces to

\[k_{eq} = k_{1} + k_{2}\]

For more springs \(k = k_{1} + k_{2} + k_{3} + \ldots\ldots\)

Series combination

When spring are connected in series then we can replace them by a single spring of spring constant \(k_{eq}\) where \(\frac{1}{k_{eq}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}\). This situation is shown in figure.

Here we present the proof. As the spring are massless, so force in the spring will be the same

\[F_{1} = F_{2} = F\]

Now for equivalent spring \(F = k_{eq}y\), as spring constants are not equal so extensions will not be equal, but total extension \(y\) can be written as sum of two extensions \(y = y_{1} + y_{2}\)
or

\[\frac{F}{k_{eq}} = \frac{F}{k_{l}} + \frac{F}{k_{2}}\ \left\lbrack \text{~}\text{as for}\text{~}F = ky,y = \frac{F}{k} \right\rbrack\]

For more than two springs Connected in series:

\[\frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}} + \ldots\]

Parts of a spring :

If a spring of force constant k of length \(l\) is cut in two parts say of \(l_{l}\) and \(l_{2}\), let us assume that new force constants are \(k_{1}\) and \(k_{2}\) for the two parts. If we connect these two parts in series, the equivalent force constant must be initial k . Thus we have

\[\frac{1}{k_{eq}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}\]

According to the molecular properties of a spring, the force constant of a part of the spring is inversely proportional to its length, which gives us

\[k_{1} = \frac{c}{l_{l}}\ \text{~}\text{and}\text{~}\ k_{2} = \frac{c}{l_{2}}\]

Where \(c\) is a positive constant depending upon the material of spring. Substituting the above values of new force constants \(k_{1}\) and \(k_{2}\) in equation, we get
or

\[\begin{matrix} & \frac{1}{k_{eq}} = \frac{l_{f}}{c} + \frac{l_{2}}{c} \\ & c = \frac{l}{k_{eq}} \end{matrix}\]

Using value of \(c\) in equation, we have

\[k_{1} = \frac{k_{eq}l}{l_{l}}\ \text{~}\text{and}\text{~}\ k_{2} = \frac{k_{eq}l}{l_{2}}\]

Spring force doesn't change instantaneously, whereas, the tension in the string changes instantaneously.
If a spring is cut, the tension in that spring becomes zero instantaneously.

Conclusion :

It is important to remember that ropes can change tension instantaneously while spring need to move to change tension, so in this example tension in spring is not changing instantaneously

Considering FBD

We know from part (a) that tension in spring is

\[T_{l}\text{~}\text{and}\text{~}\ T_{l} = 4mg\]

Writing Newton's Second Law for A

\[4mg - T_{l} = 4ma_{A}^{'}\]

Writing Newton's Second Law for \(B\)

\[T_{I} + mg - T_{2}\ ^{'} = ma_{B}\ ^{'}\]

Writing Newton's Second Law for \(C\)

\[T_{2}^{'} = 3ma_{C}^{'}\]

Quantities which may have different value from part (a)are represented using symbol ', for eg a tenstion in string \(s_{2}\) is \(T_{2}\ ^{'}\), others which have same value as part (a) have been retained with same symbol.
as \(a_{B}\ ^{'} = a_{C}\ ^{'}\)
solving we get \(a_{A}\ ^{'} = 0\ a_{B}\ ^{'} = a_{C}\ ^{'} = \frac{5}{4}\text{ }g\)

Practice Exercise

Q. 1 Find extension of spring in equilibrimum condition, all the blocks are of mass m and all springs are of spring constant k .

(i)

(ii)

(iii)

Q. 2 Two blocks A and B of mass M and 2 M respectively, are hanging from a ceiling by means of a light spring and a light string as shown. If the string between the blocks is suddenly cut, tino, the accelerations of the block A and B.

Answers

Q. 1
(i) \(\frac{Mg}{2k}\);
(ii) \(\frac{mg}{4k}\);
(iii) \(\frac{4mg}{3k}\)
Q. \(2a_{A} = 2\text{ }g;a_{B} = g\)

Motion in Accelerated Frames :

Till now we have restricted ourselves to apply Newton's laws of motion, only to describe observations that are made in an inertial frame of reference. In this part, we learn how Newton's laws can be applied by an observer in a noninertial reference frame. For example,consider a block kept on smooth surface of a compartment of train.

If the train acclerates, the block accelerates toward the back of the train. When observed from the train we may conclude based on Newton's second law \(F =\) ma that a force is acting the block to cause it to accelerate, but the Newton's second law is not applicable from this non- inertial frame. So we can not relate observed accleration with the Force acting on the block.

If we still want to use Newton's second law we need to apply a psuedo force, acting in backward direction, ie opposite to the aceleration of noninertial reference frame. This force explains the motion of block towards the back of train. The fictions force is equal to \(- \overrightarrow{ma}\), where \(\overrightarrow{a}\) is the acceleration of the non inertial reference frame. Fictitious force appears to act on an object in the same way as a real force, but real forces are always interactions between two objects, on the other hand there is no second object for a fictitious force.

\[\text{~}\text{pseudo force,}\text{~}{\overrightarrow{F}}_{P} = - m{\overrightarrow{a}}_{0}\]

where \(a_{0}\) is acceleration of non inertial reference frame
Thus, we may conclude that pseudo force is not a real force. When we draw the free body diagram of a mass, with respect to an inertial frame of reference we apply only the real forces (forces which are actually acting on the mass), but when the free body diagram is drown from a non-inertial frame of reference a pseudo force (in addition to all real forces) has to be applied to make the equation
\(\overrightarrow{F} = {ma}_{0}\), valid in this frame also.
Suppose a block A of mass \(m\) is placed on a lift ascending with an acceleration \(a_{d}\). Let \(N\) be the normal reaction between the block and the floor of the lift Free body diagram of \(A\) is shown in figure.

for observer 1 (Inertial reference frame):

For observer 2 (Non inertial reference frame):

\[\therefore\ N = mg + ma\]

From inertial frame of reference (Ground)

FBD of block w.r.t. ground (Apply real forces):
With respect to ground block is moving with an acceleration ' \(a\) '.

\[\begin{array}{r} \therefore\ \Sigma F_{y} = 0\ \Rightarrow \ Ncos\theta = mg\#(i) \end{array}\]

\[\begin{array}{r} \text{~}\text{and}\text{~}\ \Sigma F_{x} = ma\ \Rightarrow \ Nsin\theta = ma\#(ii) \end{array}\]

From Equation (i) and (ii)

\[\begin{matrix} a & \ = gtan\theta \\ F & \ = (M + m)a \\ & \ = (M + m)gtan\theta \end{matrix}\]

From non-inertial frame of reference (Wedge)

FBD of block w.r.t. wedge (real force + pseudo force)

w.r.t. wedge block is stationary

\[\begin{array}{r} \therefore\ \Sigma F_{y} = 0\ \Rightarrow \ Ncos\theta = mg\#(iii) \end{array}\]

\[\begin{array}{r} \text{~}\text{and}\text{~}\ \Sigma F_{x} = ma\ \Rightarrow \ Nsin\theta = ma\#(iv) \end{array}\]

From equation (iii) and (iv), we will get the same result

\[F = (M + m)gtan\theta\]

Frictional forces are unavoidable in our daily lives. If we were not able to counteract them, they would stop every moving object and bring to a halt every rotating shaft. About \(20\%\) of the gasoline used in an automobile is needed to counteract friction in the engine and in the drive train on the other hand, if friction were totally absent, we could not get an automobile to go anywhere, and we could not walk or ride a bicycle. We could no hold a pencil, and, if we could, it would not write. Nails and screws would be useless, woven cloth would fall apart, and knots would untie.
When surfaces slide or tend to slide over one another, a force of friction acts. Friction is caused by the irregularities in the surfaces in mutual contact, and it depends on the kinds of material and how much they are pressed together. Even surfaces that appear to be very smooth have microscopic irregularities that obstructs motion. Atoms cling together at many points of contact. When one object slides against another, it must either rise over the irregular bumps or else scrape atoms off. Either way requires force. Although the details of friction are quite complex at atomic level, it ultimately involves the elctromagnetic force between atoms & molecules.

The direction of the friction force is always in a direction opposing relative motion. An object sliding down an incline experiences friction directed up the incline; an object that slides to the right experiences friction toward the left. Thus, if an object is to move at constant velocity, a force equal to the opposing force of friction must be applied so that the two forces exactly cancel each other. The zero net force then results in zero acceleration and constant velocity.

Static Friction

Static friction acts when two contact surfaces are not moving relative to each other. For example, consider a block on a horizontal table, as in figure. If we apply an external horizontal force F to the block, acting to the right, the block remains stationary if F is not too large. The force that counteracts F and keeps the block from moving acts to the left and is the frictional force f . As long as the block is not movng, \(f = F\). Since the block is stationary, we call this frictional force the force of static friction, \(f_{s}\).
If we keep two books one on top the other and now we slowly push the lower book, both the books move together, the force moving the upper book is friction. Since the books are moving with respect to ground but they are not moving

\(F = f_{v}\), As long as the block is not moving

Two books kept on top of each other & the lower book being pushed slowly

with respect to each other, this force of friction between two books is of static nature.

Similarly when we walk on ground the friction force acting on the foot is of static nature. You may be suprised but the foot in contact with ground is not moving while the other foot is moving. During this time friction acting between foot and ground is static friction.

If we are increasing our speed then static friction is acting in forward direction. This happens because we try to pull our leg backward while walking in forward direction. This means foot is trying to move in backward direction with respect to ground and ground applies friction in forward direction

Static friction is self-adjusting and it opposes the tendency of relative motion

In fact, in the previous case of books also, we can see that static friction acting in forward direction on upper book has given it some motion so it can move with the lower book. Another point to be noted here is that even by third law, pair of static friction is opposing relative motion by trying to slow down the lower book (which is being accelerated externally).
for example in figure we assume that the block is stationary and we can see that static friction is acting in such a direction so as to oppose the relative motion. F represents external force and \(f_{s}\) represents friction.

Block is stationary due to static friction

The magnitude of the static friction between any two surfaces in contact can have the values

\[f_{s} \leq \mu F_{N}\]

where the dimensionless constant \(\mu_{s}\) is known as the coefficient of static friction and \(F_{N}\) is the magnitude of the normal contact force exerted by one surface on the other. The equality in equation holds when the surfaces are on the verge of slipping, that is when \(f_{s} = f_{s,max} = \mu_{s}F_{N}\), This maximum value of \(f_{s}\) is called limiting friction. This situation is called impending motion. The inequality hold when the surfaces are not on the verge of sliping. Maximum strength of the joints formed is directly proportional to the normal contact force, that is \(f_{s,max} \propto F_{N}\).

Maximum strength also depends on the roughness of contact surface \(f_{s,max}\left( \right.\ \) also called \(\left. \ f_{\text{limiting}\text{~}} \right) = \mu_{s}\text{ }N\). Magnitude of static friction is self-adjusting such that relative motion does not start (but still it has maximum value).

Let us say we are applying force \(F\) on a block kept on horizontal rough surface with coefficient of static friction \(\mu_{s} = 0.1\) & mass of block is 5 kg . When applied F is less than 5 N the value of static friction is
equal to the applied force, not 5 N . It is the maximum value of friction. But when applied force F is equal to 5 N , the value of static friction is 5 N

Kinetic friction acts when there is relative motion between two surfaces in contact. It acts always opposite to the relative velocity as we can see in figure. The magnitude is not self-adjusting as in static friction, it is always is equal to \(\mu_{k}F_{N}\).

Experimentally, we find that, to a good approximation, both \(f_{s,max}\) and \(f_{k}\) are proportional to the magnitude of the normal force. The following empirical laws of friction summarize the experimental observations :

1. The magnitude of the force of kinetic friction acting between two surfaces is
   \[f_{k} = \mu_{k}N \]where \(\mu_{k}\) is the coefficient of kinetic friction. Although the coefficient of kinetic friction can vary with speed, we shall usually neglect any such variations in problems.

2. The value of \(\mu_{k}\) and \(\mu_{s}\) depend on the nature of the surfaces.

3. \(\ \mu_{k}\) is generally less than \(\mu_{s}\).

The actual value depends on the degree of smoothness and other environmental factors. For example, wood may be prepared at various degrees of smoothness and the friction coefficient will vary. Dust impurities, surface oxidation etc. have a great role in determining the friction coefficient. Suppose we take two blocks of pure copper, clean them carefully to remove any oxide or dust layer at the surfaces, heat them to push out any dissolved gases and keep them in contact with each other in an evacuated chamber at a very low pressure of air. The blocks stick to each other and a large force is needed to slide on over the other. The friction coefficient as defined above, becomes much larger than one this is called cold welding. If a small amount of air is allowed to go into the chamber so that some oxidation takes place at the surface, the friction coefficient reduces to usual values.
4. The direction of the friction force on an object is parallel to the surface with which the object is in contact and opposite to the motion (kinetic friction) or the tendency of motion (static friction) of the object relative to the surface.
5. The coefficients of friction are nearly independent of the area of contact between the surfaces. We might expect that placing an object on the side having the more area might increase the friction force. While this provides more points in contact, the weight of the object is spread out over a larger area, so that the individual points are not pressed so tightly together. These effects approximately compensate for each other, so that the friction force is independent of the area. So those extra wide tires you see on some cars provide no more friction than narrower tires. The wider tire simply spreads the weight of the car over more surface area to reduce heating and wear. Similarly, the friction between a truck and the ground is the same whether the truck has four tires or eighteen! More tires spread the load over more ground area and reduces the pressure per tire. Interestingly, stopping distance when brakes are applied is not affected by the number of tires.But the wear that tires experience, very much depends on the number of tires.

Various mechanical configurations (left) and the corresponding free-body diagrams (right). The term rough here means only that the surface is not frictionless.

A block is pulled up the rough incline

Two blcoks in contact, pushed by the force F

to the right on a frictionless surface.

Two masses connected by a light cord.
The surface is rough and the pulley is frictionless

Practice Exercise

Q. 1 You press your physics textbook flat against a vertical wall with your hand. What is the direction of the friction force exerted by the wall on the book?
Q. 2 A crate is located in a truck. The truck accelerates to the east, and the crate moves with it, without sliding. What is the direction of the friction force exerted by the crate on the truck?
Q. 3 A block lies on a floor. (a) What is the magnitude of the frictional force on it from the floor? (b) If a horizontal force of 5 N is now applied to the block, but the block does not move, what is the magnitude of the frictional force on it? (c) If the maximum value of the static frictional force on the block is 10 N , will the block move if the magnitude of the horizontally applied force is 8 N ? If it is 12 N ? (d) What is the magnitude of the frictional force in part (c)?
Q. 4 A 1 kg hanging block is connected by a string over a pully to a 2 kg block sliding on a flat table. If the coefficient of sliding friction is 0.20 , find the tension in the string.

Q. 5 A 25 kg block is initially at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving with constant speed. Find the coefficients of static and kinetic friction from this information.
Q. 6 Two blocks connected by a massless rope are being dragged by a horizontal force F . Suppose that \(F = 60\text{ }N,{\text{ }m}_{1} = 12\text{ }kg,{\text{ }m}_{2} = 18\text{ }kg\), and coefficient of kinetic friction between each block and the surface is 0.10 . (a) Find the magnitude of the acceleration of the system.

Answers

Figure shows a block of weight \(mg\) resting on a rough surface. A horizontal force is applied to the block. Force P is gradually increased from zero.

When force P is very small, the block does not move form condition of equilibrium,

\[\begin{matrix} & \Sigma F_{x} = P - F_{\text{friction}\text{~}} = 0;F_{\text{friction}\text{~}} = P \\ & \Sigma{\text{ }F}_{y} = N - mg = 0;N = mg \end{matrix}\]

Friction force counter balance external force, till the block is static. This friction force is static friction. As external force is increased, static friction also increases.

As the force P is gradually increased, a limiting point is reached where friction force F is not sufficient to prevent onset of motion. When the block is about to move, the state of motion is called impending state of motion. At this point friction force has maximum value and given by the equation

\[F_{\max} = \mu_{s}\text{ }N\]

where \(\mu_{s}\) is coefficient of static friction, N is normal reaction. If force P is greater than \(F_{\max}\), the block will have a resultant froce \(P - F_{\text{max}\text{~}}\) on it. The block will accelerate in the direction of resultant force when sliding motion starts. At this moment friction force is given by

\[F = \mu_{k}N\]

Where \(\mu_{k}\) is coefficient of kinetic friction. The figure shows variation of friction force versus external force graph.

Angle of friction ( \(\phi\) )