Video Lecture

Theory For Making Notes

Newton’s Laws

Newton’s First Law

(a)

Everybody continues in its state of rest of uniform motion in a straight line unless it is compelled by a resultant force to change that state.

(b)

This law is also known as law of inertia.

(c)

Inertia is the property of inability of a body to change its position of rest or uniform motion in a straight line unless some external force acts on it.

(d)

Mass is a measure of inertia of a body.

(e)

A frame of reference in which Newton’s first law is valid is called inertia frame, i.e., if a frame of reference is at rest or in uniform motion it is called inertial, otherwise non-inertial.

Newton’s Second Law

The rate of change of momentum of a body is directly proportional to the net external force acting on it and takes place in the direction of the force.

$\displaystyle \vec{F}\,=\,\frac{{d\vec{p}}}{{dt}}\,=\,\frac{{md\vec{v}}}{{dt}}\,=\,m\vec{a}$

The net external force acting on a particle produces acceleration. The magnitude of the acceleration produced depends on the quantity of matter being acted upon. The quantity of matter is referred to as the mass. The direction of acceleration is the direction of the net external force.

So mathematically,

$\displaystyle \vec{F}_{net}$ = m $\displaystyle \vec{a}$ ,

$\displaystyle \vec{F}_{net}$ includes vector sum of all external forces.

$\displaystyle \vec{F}\,=\,m\left( {\frac{{d{{v}_{x}}}}{{dt}}\hat{i}+\frac{{d{{v}_{y}}}}{{dt}}\hat{j}\,+\,\frac{{d{{v}_{z}}}}{{dt}}\hat{k}} \right)$

$\displaystyle {{F}_{x}}\hat{i}+{{F}_{y}}\hat{j}+{{F}_{z}}\hat{k}=m\frac{{d{{v}_{x}}}}{{dt}}\hat{i}+m\frac{{d{{v}_{y}}}}{{dt}}\hat{j}+m\frac{{d{{v}_{z}}}}{{dt}}\hat{k}$

In component Form the force can be written as follows

$\displaystyle {{F}_{x}}\,=\,m\frac{{d{{v}_{x}}}}{{dt}}$

$\displaystyle {{F}_{y}}=m\frac{{d{{v}_{y}}}}{{dt}}$

$\displaystyle {{F}_{z}}=m\frac{{d{{v}_{z}}}}{{dt}}$

Similarly acceleration can also be written in component form

$\displaystyle {{a}_{x}}=\frac{{{{F}_{x}}}}{m}$

$\displaystyle {{a}_{y}}=\frac{{{{F}_{y}}}}{m}$

$\displaystyle {{a}_{z}}=\frac{{{{F}_{z}}}}{m}$

Newton’s Third Law

To every action (Force) there is an equal and opposite reaction (Force). Forces always occur in pair. Force is the result of mutual interaction between two bodies. Mutual forces between two bodies are always equal and opposite.

Let body B exerts force $\displaystyle {{\vec{F}}_{{AB}}}$ on body A, body A exerts a force $\displaystyle {{\vec{F}}_{{BA}}}$ on body B. These forces are related by $\displaystyle {{\vec{F}}_{{AB}}}=-{{\vec{F}}_{{BA}}}$

The negative sign represents opposite direction. The two forces involved in any interaction between two bodies are called action and reaction. But we cannot say that this particular force is action and the other one is reaction. Action and Reaction always acts simultaneously on different bodies.

Illustration

An object of 5 kg is far away from the surface of the earth and has an acceleration of 6 m/s² towards the center of earth. Find the force exerted by the object on the earth.

Solution

Force on object by the earth, $\displaystyle {F}_{OE}={{m}_{{ob}}}.{a}_{ob}=(5\times 6)N=30N$

Force between earth and object makes an action reaction pair thus force exerted by object on earth is also 30 N directed away for the center of earth.

Illustration

The momentum of a block is given by $\displaystyle \vec{P}=(10{{t}^{2}}\hat{i}+5t\hat{j})$ kg m/s. Find the force acting on block at t = 2 s.

Solution

$\displaystyle\vec{F}=\frac{{\vec{dP}}}{{dt}}=\frac{d}{{dt}}(10{{t}^{2}}\hat{i}+5t\ \hat{j})=20t\hat{i}+5\ \hat{j}$

At t = 2 s, $\displaystyle\vec{F}=(40\hat{i}+5\hat{j})\ \ N$

Illustration

A body of mass 6 kg moves in a straight line according to the equation

x = t³ – 75t

where x denotes the distance in metre and t the time in second. The force on the body at t = 4 second is

(a) 64 newton (b) 72 newton (c) 144 newton (d) 36 newton

Solution

x = t³ – 75 t

$\displaystyle v=\frac{{dx}}{{dt}}=3{{t}^{2}}-75$

$\displaystyle a=\frac{{dv}}{{dt}}=6t$

t = 45; a = 24 ms^{– 2}

F = ma = 6 × 24 = 144 N Ans. (c)

Basic force in nature

(i)

There are four types of basic forces that exist in nature :

(a) Gravitational force

(b) Electromagnetic force

(d) Weak force

(ii)

Many other well known forces like frictional force, elastic force, viscous force, spring force, intermolecular force are manifestations of electromagnetic forces.

(iii)

The gravitational and the electromagnetic forces are long range forces having infinite range while nuclear and weak forces are very short range forces.

(iv)

The nuclear force is the strongest force while the gravitational force is the weakest force of nature. Electromagnetic force is about 10³⁶ times stronger than gravitational force while weak forces are about 10²⁵ times as strong as gravitational force. The relative magnitudes of different forces can be expressed as :

(v)

$\displaystyle {{F}_{G}}:\,{{F}_{W}}:{{F}_{E}}:{{F}_{N}}\,::1:{{10}^{{25}}}:{{10}^{{36}}}:{{10}^{{38}}}$

i.e., $\displaystyle {{F}_{N}}>\,\,{{F}_{E}}>{{F}_{W}}>{{F}_{G}}$

(vi)

Gravitational and electromagnetic forces are central or conservative forces and obey inverse square law, while nuclear force is a non-central force and varies inversely with some higher power of distance.

(vii)

There are certain reactions like beta-decay, which can be explained only on the basis of weak forces.

(viii)

Electromagnetic interactions can be explained on the basis of exchange of photons; gravitational force on the basis of exchange of graviton; weak forces on the basis of exchange of $\displaystyle {{W}^{+}},\,\,{{W}^{-}},\,\,{{Z}^{0}}$ bosons and nuclear forces on the basis of exchange of $\displaystyle \pi$ -mesons.

Force

We know by experience that all bodies in nature interact in some way with one another. Force is a measure of the interaction between the bodies. The force may either produce deformation (change in the size or shape of bodies) or acceleration (change in their velocity).

It can be described as a push or a pull. The pull is always oriented away from the body; and the push is always oriented towards the body.

Force is a vector quantity. Every force has a definite direction and the result of its action depends on the direction and the magnitude of the force.If several forces are applied to a particle, they can be replaced by the resultant force.

Units of force

Absolute units : (i) Newton (S.I.) (ii) Dyne (C.G.S)

Gravitational units : (i) Kilogram-force (M.K.S.) (ii) Gram-force (C.G.S)

Newton : One Newton is that force which produces an acceleration of $1\,m/{{s}^{2}}$ in a body of mass 1 Kilogram.

$\therefore 1$ Newton $=1kg-\,m/{{s}^{2}}$

Dyne : One dyne is that force which produces an acceleration of $1cm/{{s}^{2}}$ in a body of mass 1 gram.

1 Dyne $=1gm\,cm/{{\sec }^{2}}$

Relation between absolute units of force 1Newton $={{10}^{5}}$ Dyne

Kilogram-force :It is that force which produces an acceleration of $9.8m/{{s}^{2}}$ in a body of mass 1 kg.

1 kg-f = 9.80 Newton

Gram-force : It is that force which produces an acceleration of $\displaystyle 980cm/{{s}^{2}}$ in a body of mass 1gm.

1 gm-f = 980 Dyne

Free body Diagram

In mechanics, diagram of an isolated body along with forces acting on it is very useful in the analysis of mechanical problems. Such diagram is defined as free-body diagram.

Some Common Forces

There are, basically, five forces which we commonly encounter in mechanics, namely the weight, the normal forces, the friction force, the tension, the spring force

Weight

Weight is the force with which the earth attracts every other body. It is also called the force of gravity or the gravitational force. If they do not fall to the earth, then their motion is restricted by certain other bodies : a support, string, spring, etc. Bodies that restrict the motion of other bodies are called constraints. These bodies restrict the motion of the given bodies. For example, the surface of the table is the constraint for all objects lying on it, the floor serves as the constraint for the table, etc.

Normal force

It is the reaction force provided by the surface in contact with the body against the component of force by which body press the surface.

It is a self-adjusting force act on the body in a direction normal to the surface through the points of contacts and with a pushing tendency that is never being achieved by it.

Normal reaction is not a basic fundamental force rather it arises due to the electromagnetic interaction between the atoms of body and of surface which are in contact with each other. Its magnitude can be calculated using Newton’s Law of motions as follows:

Consider a block of mass m placed at rest on the horizontal surface may be smooth or rough as shown in figure. The block experience two external forces:

One is the weight mg arises due to the gravitational interaction with earth.
Second force N arises due to the interaction of block with surface called normal reaction as shown in the F.B.D. of block with respect to ground observer.

As the acceleration of the block is zero in the frame of reference of ground observer, the second law of motion suggest that

$N-mg=0$ …. (i)

or $N=mg$

Thus the force exerted by ground on the block is equal to the weight of block itself. Further force exerted by block on the ground is N in downward direction too as suggested by Newton’s third law. Hence, the normal reaction exerted by the ground is equal and opposite to the force exerted by block on ground which is equal to the weight of the block in this situation.

If the surface is accelerating upward with an acceleration a then, in the reference of the observer on the ground, the equation (i) modified as

$N-mg=ma$

$N=m(g+a)$ …. (ii)

Thus, the normal reaction by the ground on block is increase by an amount ma. It is obvious because now the surface not only holding a block to fall down but also giving it an upward push to accelerate upward.

The normal reaction provided by surface to stationary block in three commonly encountered situations as shown in figure, is given as:

A block of mass m placed on ground and is pulled with a force F inclined at an angle $\theta$ with horizontal.

From the Newton’s Laws, the equation in Y-direction gives

$N-mg+F\sin \theta =0$

or $N=mg-F\sin \theta$ , The same equation can also be obtained by equating the upward and downward forces acting on the block

Similarly, if the block is pushed by force F, at an angle $\theta$ with horizon, then

$N=mg+F\sin \theta$

A block of mass m placed on an inclined ground of inclination $\theta$ .

In this case it is always better to find the components of the forcs along and perpendicular to the inclined surface

The equation of motion in Y–axis gives

$mg\cos \theta -N=0$ or $N=mg\cos \theta$

A block of mass m is pushed against a vertical wall by force F

The equation of motion in X-direction gives

$F-N=0$ or $N=F$

Note that the normal force does not depend on the weight of body but generally a part of weight press surface, so N contains a term of weight.

Frictional Forces

The component of the contact force parallel to the contact surface is called the frictional force. The direction of the frictional force is opposite to the relative motion (or attempted motion) of the two surfaces in contact.

Spring Force

When a spring is deformed(stretched or compressed), it resists the deformation by generating a force called the spring force.

The magnitude of spring force ${{F}_{s}}$ is proportional to the deformation `x’ as

${{F}_{s}}=kx$

Where k is called the stiffness constant or spring constant. In the vector form, it may be written as

${{\vec{F}}_{s}}=-k\vec{x}$

The negative sign indicates that the spring force is always opposite to the direction of deformation.

F_ext = external force applied on the force, F_S = spring force resisting the deformation

Internal and External force :

The forces of interaction between portions of a system of bodies being considered are called internal forces. The force exerted on bodies of a given system by bodies not included in this system are called external forces. A system of bodies on each of which no external forces act is called a closed (isolated) system.

Tension

It is again a self-adjusting reaction force provided by the rope, string, chain etc. against the pulling forces acting at its ends.

It is again consequence of the electromagnetic interaction between the atoms of the string, rope or chain etc.

When the two equal forces at its two ends pull rope, the rope remains stationary but it become taut. The extreme atom, at which external force is acting, is kept stationary by a force exerted by the second inner atom. As this is an interaction force, so to the first atom also exert pulling force on second atom that is balanced by the force exerted on it by third atom. This pulling between the atoms extended to the whole of the rope and make the rope stretched but with zero net force at each point of rope.

In figure, a rope of three atoms pulled by two equal forces of magnitude F each is shown in figure. It is clear from figure that tension at each point of the rope is same and equal to

$T=F$

When rope is pulled with unequal forces ${{F}_{1}}$ and ${{F}_{2}}$ , then tension is not same at each point of the rope. It accelerate with acceleration a given by

$a=\frac{{{{F}_{1}}-{{F}_{2}}}}{m}$ for ${{F}_{1}}>{{F}_{2}}$

Let T be the tension at point having a distance x from the end A, then for differential element at x equation of motion suggest

$(T+dT)-T=(dm)a$

or $dT=dma$

If the mass in rope is distributed uniformly, then $dm=\frac{m}{L}dx$ and hence

$\frac{{dT}}{{dx}}=\frac{m}{L}\left( {\frac{{{{F}_{1}}-{{F}_{2}}}}{m}} \right)$ or $-\frac{{dT}}{{dx}}=\frac{{{{F}_{1}}-{{F}_{2}}}}{L}=\mathbf{Constant}$

Here –ve sign shows that tension decreases from ${{F}_{1}}$ to ${{F}_{2}}$ along string. Since the tension at point A is equal to ${{F}_{1}}$ , so the tension at x is

$T={{F}_{1}}-\frac{{{{F}_{1}}-{{F}_{2}}}}{L}\,x$

Thus, the variation of T with x is shown in figure.

Note that if the rope is massless, then the tension at each point of the rope is same even if it is pulled with unequal forces.

Illustration

A block of mass m is placed on a rough inclined surface of inclination $\theta$ as shown in figure.Draw the F.B.D. of block

Solution

The F.B.D. of the block is, shown in figure, with N as the normal reaction and f is the friction force.

Illustration

A system of block and wedge is shown in figurein equilibrium. Show the F.B.D of Mass M and ball m.

Solution:

The F.B.D. of block and wedge in the frame of wedge are shown in figure

Equilibrium of Concurrent Force

(1)

If all the forces working on a body are acting on the same point, then they are said to be concurrent.

(2)

A body, under the action of concurrent forces, is said to be in equilibrium, when there is no change in the state of rest or of uniform motion along a straight line.

(3)

The necessary condition for the equilibrium of a body under the action of concurrent forces is that the vector sum of all the forces acting on the body must be zero.

(4)+

Mathematically for equilibrium ${\vec{F}_{net}}$ = 0

or individual components along x , y and z axis should be zero

Hence $\sum{{{{F}_{x}}=0}}$ ;

$\sum{{{{F}_{y}}=0}}$ ;

$\sum{{{{F}_{z}}=0}}$

(5)

Three concurrent forces will be in equilibrium, if they can be represented completely by three sides of a triangle taken in order.

(6) Lami’sTheorem : For three concurrent forces in equilibrium

$\frac{{{{F}_{1}}}}{{\sin \alpha }}=\frac{{{{F}_{2}}}}{{\sin \beta }}=\frac{{{{F}_{3}}}}{{\sin \gamma }}$

Illustration

A ball of mass 5kg is suspended with two strings as shown in figure.

(a) Draw the F.B.D. of the block and determine the tension in both strings.

(b) If the string (1) is cut down suddenly, find the tension in the string (2) at this instant (say A).

(c) What is the difference between the state of rest of ball at this instant before and after the string (1) is cut down?

Solution:

Before the string is cut

The ball, before the string is cut, is at rest and is in an equilibrium state i.e., the ball has zero velocity as well as acceleration along all directions. The F.B.D of the ball is shown in figure. The equations of motion of ball are:

Along X-axis

${{T}_{1}}={{T}_{2}}\sin 60{}^\circ$ ….. (i)

Along Y-axis

${{T}_{2}}\cos 60{}^\circ =mg$ ….. (ii)

From (i) and (ii), we get

${{T}_{1}}=mg\tan 60{}^\circ$ or ${{T}_{1}}=5\times 10\times \sqrt{3}=50\sqrt{3}\,\mathbf{N}$

and ${{T}_{2}}=mg\sec 60{}^\circ$ or ${{T}_{2}}=5\times 10\times 2=100\,\mathbf{N}$

After the string is cut

When the string is cut down, the ball does not remain in an equilibrium state along the direction perpendicular to the string as suggested by the F.B.D. of the block.

The equation of motion of blocks suggested that

Along Y-axis

$-mg\sin 60{}^\circ =m{{a}_{y}}$ …(iii)

Along X-axis

$T_{2}^{‘}-mg\cos 60{}^\circ =0$ $(\because \,\,\,\,{{a}_{x}}=0)$ …(iv)

From (iii) and (iv), we get

${{a}_{y}}=g\sin 60{}^\circ$

and $T_{2}^{‘}=mg\cos 60{}^\circ$ or $T_{2}^{‘}=5\times 10\times \frac{1}{2}=25\,\,\mathbf{N}$ .

Illustration

A block A is kept stationary on a wedge B by applying a force F as shown in figure.

(a)Draw the F.B.D. of the block.

(b) Determine the normal reaction exerted on block by the wedge.

Assume that there is no friction anywhere in the system.

Solution:

(a) The F.B.D. of block A is shown in figure.

(b) The equations of motion of block are:

Along X-axis

$F-N\sin \theta =m{{a}_{x}}$ and ${{a}_{x}}=0$

$F=\mathbf{N}\sin \theta$ …(i)

Along Y-axis

$N\cos \theta -mg=m{{a}_{y}}$ and ${{a}_{y}}=0$

$N\cos \theta =mg$ …(ii)

Thus the normal reaction exerted on block by the wedge is

N=F/Sinθ … (iii)

F=mg tanθ …….(iv)

Illustration

A block of mass M is hung via massless string, pulley and spring system as shown in figure.

(a) Determine the tension in string and the extension of spring, if stiffness constant of spring is k.

(b) If the pulley also has mass M, then how much the block has shifted from its position shown in figure now.

Solution:

(a) Let T is the tension in string and is the extension of spring from its natural length in a state of equilibrium. The F.B.D. of the block and pulley is shown in figures and suggest that:

For block: $T=Mg$ …(i)

For Pulley: $2T=k{{x}_{o}}$ …(ii)

From (i) and (ii) ${{x}_{o}}=\frac{{4Mg}}{k}$ ….(iii)

(b) Let x be the extension of spring when pulley also has mass M. Then the F. B.D. of the pulley suggest that

$kx=2T+Mg$ or $kx=k{{x}_{o}}+Mg$ or $x=\frac{{5Mg}}{k}$ …(iv)

From (iii) and (iv), shift in the position of the pulley is

$\Delta x=x-{{x}_{o}}$ $=\frac{{5Mg}}{k}-\frac{{4Mg}}{k}=\frac{{Mg}}{k}$ …(v)

Since shift in the position of pulley is $\Delta x$ downward, so it releases the string by a length $2\Delta x$ . As one end of the string is fixed, so the second end where block is attached fall by a distance $2\Delta x$ to keep the string taut. Thus, shift in the position of block in downward direction is

$\Delta x’=\frac{{2Mg}}{k}$ …(vi)

Illustration

Two blocks A and B, of masses 2.9 kg and 1.9 kg respectively, are suspended from a rigid support by inextensible strings. The length of each string is one metre and the upper string has negligible mass while the lower string has a uniform mass of 0.2 kg. Determine the tension at the mid points of strings when the support is

(a) stationary

(b) accelerating upward at the rate of $\mathbf{0}\mathbf{.2}\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

Solution:

The problem can be solved by assuming that block A has mass (2.9+0.1)kg and block B has mass (0.9+0.1) kg and both the blocks are now connected with a massless string. The tension in the string connecting A and B is now same at each point and its value is equal to the tension at the mid point of a massive rope of mass 0.2 kg.

(a) When the support is stationary, then the tension in lower string balance the weight of block B and the tension in upper string balance the weight of both blocks A and B. Thus,

${{T}_{{lower}}}={{m}_{B}}g=2g=19.6\,\mathbf{N}$

and ${{T}_{{upper}}}=({{m}_{B}}+{{m}_{A}})g=(2+3)g=49\mathbf{N}$

(b) When the support is acceleration upward, then, in the frame of reference of support, the system is in equilibrium due to the downward pseudo force, which increases the effective value of g by $\mathbf{0}\mathbf{.2}\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$ .

Thus, the effective value of g is

${{g}_{{eff}}}=9.8+0.2=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

Now, the tension in strings is

${{T}_{{lower}}}={{m}_{B}}{{g}_{{eff}}}=2\times 10=20\mathbf{N}$

and ${{T}_{{upper}}}=({{m}_{B}}+{{m}_{A}}){{g}_{{off}}}=5\times 10=50\mathbf{N}$

(c) When the support is acceleration downward, the effective value of g decreases by an amount of $\mathbf{0}\mathbf{.2}\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$ .

Now, ${{g}_{{eff}}}=9.8-0.2=9.6\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

and ${{T}_{{lower}}}=2\times 9.6=19.2\,\mathbf{N}$

${{T}_{{upper}}}=5\times 9.6=48.0\,\mathbf{N}$

Illustration

A block of mass 25kg is raised by a 50 kg man in two different ways as shown in figure.

(a) What is the action on the floor by man in the two cases?

(b) If floor yields to a normal force of 700 N, which mode should the man adopt to lift block without the floor yield?

Solution:

(a) To raise the 25kg block upward, tension T required in the string is 25g. If N is the normal reaction exerted by floor on man, then the F. B.D. of man in two situations become as:

For case (A):

$T+mg={{N}_{1}}$

or ${{N}_{1}}=25g+50g=75g=750\,\mathbf{N}$

For case (B)

${{N}_{2}}=mg-T$

or ${{N}_{2}}=50g-25g=25g=250\mathbf{N}$

Since ${{N}_{1}}>700\,\mathbf{N}$ and ${{N}_{2}}<700\,\mathbf{N}$ , so mode (B) should be adopted by the man to lift up block without getting yield the floor.

Practice Questions (Basic Level)

A homogeneous rod of length L is acted upon by two forces F₁ and F₂ applied to its ends and directed opposite to each other. Find the magnitude of the pulling force F at a distance l from the end where F₁ is applied.

(a) $\frac{{\left( {{{F}_{2}}-{{F}_{1}}} \right)l}}{L}+{{F}_{2}}$

(b) $\frac{{\left( {{{F}_{2}}+{{F}_{1}}} \right)l}}{L}+{{F}_{1}}$

(d) $\frac{{\left( {{{F}_{2}}-{{F}_{1}}} \right)l}}{L}-{{F}_{1}}$

Ans (c)

A particle of mass 10^-2 kg is moving along the positive x-axis under the influence of a force F(x) = $\frac{k}{{2{{x}^{2}}}}$ where k = 10^-2 Nm². At time t = 0 it is at x = 1.0 m and its velocity is v = 0. Its velocity and time at x=2m is

(a) 1.0 m/s , t = 1.48 s

(b) 0.707 m/s , t = 5.65 s

(d) 2.0 m/s , t = 1.20 s

Ans (b)

Figure shows the position–time graph of a particle of mass 4 kg in one–dimensional motion. For t< 0, t> 4s, 0 <t< 4s choose the correct statement

(a) Particle is at rest for t<0 and t>4s

(b) Particle is in uniform motion for 0<t<4

(d) All of the above statements are correct

Ans (d)

A man of mass 70 kg stands on a weighing scale in a lift which is moving. What would be the readings on the scale in each case?

(i) Lift is moving upwards with a uniform speed of 10 ms^–1

(ii) Lift is moving downwards with a uniform acceleration of 5 ms^–2

(iii) Lift is moving upwards with a uniform acceleration of 5 ms^–2

(iv) If the lift mechanism failed and it falls freely under gravity

Choose the correct option for all above given situations

(a) Reading is 35 kg

(b) Reading is 70 kg

(d) Reading is 105 kg

Ans (i) -(b), (ii) – (a), (iii) – (d), (iv) – (c)

A horizontal force of 600 N pulls two masses 10 kg and 20 kg (lying on a frictionless table) connected by a light string. What is the acceleration and tension in the string? Does the answer depend on which mass end the pull is applied?

(a) a = 20 ms^–2, T = 400 N, answer depend on which mass pull is applied

(b) a = 20 ms^–2, T = 100 N, answer does not depend on which mass pull is applied

(d) a = 20 ms^–2, T = 350 N,answer depend on which mass pull is applied

Ans (a)

For the arrangement shown in the figure, Find the acceleration of each block with Tension in each string

(a) 2 m/s² , T₁ = 2 N , T₂ =1 N

(b) 2 m/s² , T₁ = 16 N , T₂ =10 N

(d) 10 m/s², T1 = 15 N; T2 =5 N

Ans (b)

A system consists of two blocks m₁ = 2kg and m₂ (unknown) connected by a massless rope is placed on a horizontal frictionless surface . A horizontal force of 20N acts on m₁.

(i) For what value of m₂ will the acceleration of the system be 4 m/s²

(ii) For what value of m₂ will the tension in the rope be 8 N ?

Choose the correct option

(a) (i) – 1 kg , (ii) – 5/3 kg

(b) (i) – 8 kg , (ii) – 8/3 kg

(d) (i) – 3 kg , (ii) – 4/3 kg

Ans (d)

Calculate the tension in the string shown in figure. The pulleys and the string are light and all the surfaces are frictionless. Take g = 10 m/s²

(a) 5 N

(b)10 N

(d) 50 N

Ans (a)

Two blocks with masses m₁ = 0.2 kg and m₂ = 0.3 kg hang one under the other, as shown in the figure. Find the tensions in the (massless) strings in the following situations:

(i).

The blocks are at rest.

(a) 2 N; 3 N

(b) 9 N; 8 N

(d) 5 N; 3 N

Ans (d)

(ii).

They move upward at 5 m/s.

(a) 2 N; 3 N

(b) 9 N; 8 N

(d) 7 N; 3 N

Ans(c)

(iii).

They accelerate upward at 2 m/s².

(a) 2 N; 0.4 N

(b) 6 N; 3.6 N

(d) 10 N; 10.0 N

Ans (b)

(iv).

They accelerate downward at 2 m/s².

(a) 4 N, 2.4 N

(b) 4 N, 2.0 N

(d) 4.2 N, 2.4 N

Ans (a)

(v).

If the maximum allowable tension is 10 N, what is the maximum possible upward acceleration?

(a)5 m/s²

(b)10 m/s²

(d)30 m/s²

Ans (b)

10.

A person of mass 75 kg stands on a scale in an elevator. What can you infer about the motion

(i) if the scale reads 750N

(ii) if the scale reads 600N

(iii) if the scale reads 900N

Choose the correct option for the above given situations

(a) variable acceleration

(b) v = constant

(c)acceleration of lift is 2 m/s² upward

(d)acceleration of lift is 2 m/s² downward

Ans (i) – (b), (ii) – (d) , (iii) – (c) ,

11.

A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

(a) Mode (a) Should Be Adopted

(b) Mode (b) Should Be Adopted

(d) Neither (a) Nor (b) Should Be Adopted

Ans (b)

12.

A metal sphere is hung by a string fixed to a wall. The force acting on the sphere is shown in figure. Which of the following statement is incorrect?

(a) $\vec{R}+\vec{T}+\vec{W}=0$

(b) ${{T}^{2}}={{R}^{2}}+{{W}^{2}}$

(d) $R=W\tan \theta$

Ans. (c)

13.

A fireman wants to slide down a rope. The breaking load for the rope is $\frac{3}{4}th$ of the weight of the fireman. The acceleration of the fireman to prevent the rope from breaking will be (Acceleration due to gravity is g)

(a) g/4

(b) g / 2

(d) Zero

Ans. (a)

Practice Questions (JEE Main Level)

A particle of mass 100 g starts from a point (0, –2) m with a velocity of $\vec{v}=(5\hat{i}-2\hat{j})$ m/s and moves in x-y plane under the action of force, $\vec{F}=(3\hat{i}+\hat{j})N$ . The y co-ordinate of the particle, when x co-ordinate is 10 m, will be

(a) 0 m

(b) 1 m

(d) $\frac{{10}}{9}$ m

Ans. (c)

Comprehension Based Question (2 and 3)

A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 m/s². What is the action of the block on the floor Take g = 10 m/s². Identify the action reaction pairs in the problem.

Before the floor yields

(a)20 N, vertically Upward

(b)10 N, vertically downwards

(d) 30 N, vertically downwards

Ans (a)

After the floor yields?

(a)267.3 N, vertically downward.

(b)267.3 N, vertically Upward.

(d) 272.7 N, vertically Upward

Ans (a)

Three blocks with masses m₁ = 4.5 kg, m₂ = 1.2 kg and m₃ = 2.8 kg are connected with two ropes, as shown in the figure. The horizontal surface is frictionless. Find the acceleration of the blocks and the tensions in the ropes

(a) 3m/s²,T_A= 30 N , T_B= 33.6 N

(b) 9m/s²,T_A= 20 N , T_B= 36 N

(d) 2m/s²,T_A= 36 N , T_B= 33.6 N

Ans (d)

Comprehension Based Question (5 to 7)

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 at 2 m/s² by an external force $\displaystyle {{F}_{o}}$

What is $\displaystyle {{F}_{o}}$ ?

(a) 120 N

(b) 150 N

(d) 500 N

Ans (a)

What is the net force on the rope ?

(a) 10 N

(b) 4 N

(d)5 N

Ans (b)

What is the tension at the middle of the rope ?

(a) 40 N

(b) 43 N

(d) 48 N

Ans (d)

Two blocks of masses m₁ = 2 kg and m₂ = 5 kg hang over a massless pulley A force $\displaystyle {{F}_{o}}$ = 120 N acting at the axis of the pulley accelerates the system upward. Find

the acceleration of both the masses

(a) $\displaystyle {{a}_{{2kg}}}=2m{{s}^{{-2}}}upward,{{a}_{{5kg}}}=15m{{s}^{{-2}}}downward$

(b) $\displaystyle {{a}_{{2kg}}}=30/7m{{s}^{{-2}}}upward,{{a}_{{5kg}}}=30/7m{{s}^{{-2}}}downward$

(d) $\displaystyle {{a}_{{2kg}}}=20m{{s}^{{-2}}}upward,{{a}_{{5kg}}}=2m{{s}^{{-2}}}upward$

Ans (d)

Two blocks of masses m₁ = 5 kg and m₂ = 6 kg are on either side of the wedge as shown in figure. Find the acceleration of the blocks and the force acting on the pulley must be (Ignore friction in the pulley)

(a) a= 2.45 m/s²; T = 35.25 N

(b) a= 3.45 m/s²; T = 36.50 N

(d) a= 2.45 m/s²; T = 37.25 N

Ans (c)

10.

A painter of mass M = 75 kg stands on a platform of mass m = 15 kg. He pulls on a rope that passes around a pulley, as shown in the figure. Find the pulling force in different cases

(i) He is at rest.

(a)200 N

(b)400 N

(d) 450 N

Ans (d)

(ii) He accelerates upward at 0.4 m/s².

(a) 450 N

(b) 460 N

(d) 470 N

Ans (c)

(iii) If the maximum tension the rope can withstand is 700 N, what happens when he ties the rope to a hook on the wall ?

(a) The rope will break

(b) The rope doesn’t breaks

(d) Hook will break

Ans (a)

11.

Two blocks with masses m₁ and m₂ are connected by a rope as in figure. The horizontal surface is frictionless and m₂ is subject to a horizontal force F. When F = 22 N, m₁ accelerates at 1 m/s² downward. When F = 44 N, m₁ accelerates at 1.75 m/s² upward. Determine m₁ and m₂.

(a) m₁ = 5 kg; m₂ = 3 kg

(b) m₁ = 3 kg; m₂ = 10 kg

(d) m₁ = 3 kg; m₂ = 15 kg

Ans (c)

12.

A cart with a mass M = ½ kg is connected by a string to a mass m = 200 g. At the initial moment the cart moves to the left along a horizontal plane at a speed v_o= 7 m/s. Find the magnitude and direction of velocity of the cart, the place where it will be and the distance covered by it in t = 5 s. Assume frictionless surface. Take g = 9.8 m/s².

(a)v = 7.0 m/s; d = 17.5 m

(b)v = 7.0 m/s; d = 20.0 m

(c)v = 7.0 m/s; d = 16.5 m

(d)v = 7.0 m/s; d = 10.0 m

Ans (a)

13.

Two blocks of masses 3 kg and 5 kg hang over a pulley, as shown in the figure. The 5 kg block is initially held 4 m above the floor and then released. What is the maximum height reached by the 3 kg block.

(a) 0 m

(b) 3 m

(d) 5 m

Ans (d)

14.

The system shown in the figure is in equilibrium. Find the magnitude of tension in each string: T₁, T₂, T₃ and T₄.

(a) T₁ = $\frac{{100}}{{\sqrt{3}}}$ N: T₂ = $\frac{{200}}{{\sqrt{3}}}$ N :T₃ = $\frac{{200}}{{\sqrt{3}}}$ N : T₄ = 150 N

(b) T₁ = $\frac{{100}}{{\sqrt{3}}}$ N: T₂ = $\frac{{200}}{{\sqrt{3}}}$ N :T₃ = $\frac{{200}}{{\sqrt{3}}}$ N : T₄ = 200 N

(d) T₁ = $\frac{{100}}{{\sqrt{3}}}$ N: T₂ = $\frac{{200}}{{\sqrt{3}}}$ N :T₃ = $\frac{{200}}{{\sqrt{3}}}$ N : T₄ = 250 N

Ans (b)

15.

A body moves with velocity $\displaystyle v=\text{ln}(x)\text{ m}{{\text{s}}^{{-1}}}$ , where x is its position in metre. The net force acting on body is zero at x equal to

(a) 0 m

(b) e²m

(d) 1 m

Ans. (d)

16.

The acceleration of blocks m₁ and m₂ with respect to pulley in the condition shown in figure will be (Pulley and string are ideal. g = 10 m/s²)

(a) 10 m/s²

(b) 5 m/s²

(d) zero

Ans. (b)

17.

A body of mass 2 kg has an initial velocity of 4 m/s along x axis and it is subjected to a force of 6 N in a direction perpendicular to x axis. The displacement of the body from initial point after 2 s will be

(a) 2 m

(b) 6 m

(d) 10 m

Ans. (d)

18.

A balloon of weight W newton descends with an acceleration f . The weight that must be thrown out in order to give equal upward acceleration to the Balloon will be

(a) $\frac{{Wf}}{g}$

(b) $\frac{{2Wf}}{g}$

(d) $\frac{{W(g+f)}}{f}$

Ans. (c)

19.

A man of mass 2m is pulling up a block of mass m with constant velocity. The acceleration of man is (neglect any friction).

(a) g

(b) 2g

(d) $\frac{g}{2}$

Ans. (d)

20.

Two masses m and M are connected by a light string passing over a smooth pulley. When set free m moves up by 1.4 m in 2 s. The ratio $\frac{m}{M}$ is (g = 9.8 ms–2)

(a) $\frac{13}15}$

(b) $\frac{15}{13}$

(d) $\frac{7}{9}$

Ans. (a)

21.

A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet of water at a rate of 1 kgs^–1 at a speed of 5 ms^–1. The initial acceleration of the block is

(a) $\frac{2}{5}$ ms^–2

(b) $\frac{5}{2}$ ms^–2

(d) $\frac{1}{5}$ ms^–2

Ans. (b)

Practice Questions (JEE Advance Level)

Two particles, each of mass m, are connected by a light string of length 2 L, as shown in figure. A steady force F is applied at the midpoint of the string (x = 0) at right angle to the initial position of the string. Then the acceleration of each mass in the direction at 90° to F is given by where x is the perpendicular distance of one of the particles from the line of action of F

(a) $\frac{F}{{2m}}\,\frac{x}{{{{{({{L}^{3}}\,-\,{{x}^{-2}})}}^{{1/2}}}}}$

(b) $\frac{F}{{2m}}\,\frac{x}{{{{{({{L}^{2}}\,-\,{{x}^{3}})}}^{{2/3}}}}}$

(d) $\frac{F}{{2m}}\,\frac{x}{{{{{({{L}^{-2}}\,-\,{{x}^{2}})}}^{{5/3}}}}}$

Ans. (c)

The system shown in the figure is in equilibrium. Find the tension in each string :T₁, T₂, T₃, T₄ and T₅.

(a) T₁= T₂= $\frac{{100}}{{\sqrt{3}}}$ ; T₄ = T₅ =200 N;T₃ = $\frac{{200}}{{\sqrt{8}}}$

(b) T₁= T₂= $\frac{{200}}{{\sqrt{3}}}$ ; T₄ = T₅ =100 N;T₃ = $\frac{{200}}{{\sqrt{5}}}$

(d) T₁= T₂ = $\frac{{200}}{{\sqrt{2}}}$ ; T₄ = T₅ =200 N;T₃ = $\frac{{500}}{{\sqrt{3}}}$

Ans (c)

A particle is at rest at x = a. A force $\vec{F}=-\frac{b}{{{{x}^{2}}}}\hat{i},$ directed towards the origin, begins to act on the particle. The particle starts its motion, towards the origin, along x-axis. The velocity of the particle, when it reaches at distance x from the origin, is

(a) $\sqrt{{\frac{{2b}}{m}\left( {\frac{{a-x}}{{ax}}} \right)}}$

(b) $\frac{m}{{2b}}\sqrt{{\frac{{a+x}}{{ax}}}}$

(d) infinity

Ans. (a)

A cylinder and a wedge each of mass m are touching each other. Both are free to move on smooth inclined surfaces of two fixed inclined planes. The normal exerted by the wedge on the cylinder will be

(a) $2mg\tan \alpha$

(b) $mg\tan \alpha$

(d) $mg\cos \alpha$

Ans. (b)

An object of mass m has a speed v as it passes through the origin on its way out along with positive x-axis. The object is subjected to a retarding force given by F = – Ax (A > 0). The x co-ordinate of the object when it stops is

(a) $v\sqrt{{\frac{m}{A}}}$

(b) $v\sqrt{{\frac{{2m}}{A}}}$

(d) $2v\sqrt{{\frac{m}{A}}}$

Ans. (a)

A block of mass m slides down a smooth wedge of inclination a placed on a horizontal smooth surface, the horizontal force required to keep the wedge stationary is

(a) $mg\cos \alpha$

(b) $mg\sin \alpha$

(d) $mg\cos \alpha \sin \alpha$

Ans. (d)

The 50 kg homogeneous smooth sphere rests on the 300 incline A and bears against the smooth vertical wall B. The contact force at B is (g = 10 m/s²)

(a) 250 N

(b) zero

(d) 500 N

Ans. (c)

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Theory For Notes Making

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After the string is cut

When the string is cut down, the ball does not remain in an equilibrium state along the direction perpendicular to the string as suggested by the F.B.D. of the block.

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