Laws of Motion

57 Pages·2016·1.16 MB·English

1
Concept Notes on Laws of Motion
for NEET
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5
Laws of Motion
ARISTOTLE’S FALLACY The SI unit of force is newton. (One newton force is that much
force which produces an acceleration of 1ms–2 in a body of mass
According to Aristotelian law an external force is required to keep
a body in motion. However an external force is required to 1 kg.
overcome the frictional forces in case of solids and viscous forces The CGS unit of force is dyne. (1N = 105 dyne)
in fluids which are always present in nature. The gravitational unit of force is kg-wt (kg-f) or g-wt (g-f)
1 kg-wt (kg-f) = 9.8 N, 1 g-wt (g-f) = 980dyne
LINEAR MOMENTUM (p)
Linear momentum of a body is the quantity of motion contained
Third law : To every action there is an equal and opposite
in the body. Momentum pr =mvr
reaction. For example – walking , swimming , a horse pulling a
It is a vector quantity having the same direction as the direction cart etc.
of the velocity. Its SI unit is kg ms–1. r r
F = –F
NEWTON’S LAWS OF MOTION AB BA
Action and reaction act on different bodies and hence cannot
First law : A body continues to be in a state of rest or of uniform
balance each other. Action and reaction occur simultaneously.
motion, unless it is acted upon by some external force to change
Forces always occur in pairs.
its state.
EQUILIBRIUM OF A PARTICLE
Newton’s first law gives the qualitative definition of force according
A body is said to be in equilibrium when no net force acts on the
to which force is that external cause which tends to change or
body.
actually changes the state of rest or motion of a body.
Newton’s first law of motion is the same as law of inertia given by i.e., SFr = 0
Galileo.
Then SF =0, SF =0 and SF =0
Inertia is the inherent property of all bodies because of which x y z
they cannot change their state of rest or of uniform motion unless Stable equilibrium : If a body is slightly displaced from equilbrium
acted upon by an external force. position, it has the tendency to regain its original position, it is
Second law : The rate of change of momentum of a body is directly said to be in stable equilibrium.
proportional to the external force applied on it and the change
takes place in the direction of force applied. æd2u ö
In this case, P.E. is minimum. ç =+ve÷
i.e., Fr = dpr = mdvr =mar çèdr2 ÷ø
dt dt So, the centre of gravity is lowest.
This is the equation of motion of constant mass system. For Unstable equilibrium : If a body, after being displaced from the
variable mass system such as rocket propulsion equilibrium position, moves in the direction of displacement, it is
r d(mvr) said to be in unstable equilibrium.
F =
dt æd2u ö
In this case, P.E. is maximum. ç =-ve÷
And, Fr = m(dvr)+vrdm çèdr2 ÷ø
dt dt So, the centre of gravity is highest.
130 Physics
Neutral equilibrium : If a body, after being slightly displaced
from the equilibrium position has no tendency to come back or to (cid:167)d2u (cid:183)
In this case, P.E. is constant (cid:168) (cid:32)constant(cid:184)
move in the direction of displacement the equilibrium is known to (cid:168)(cid:169)dr2 (cid:184)(cid:185)
be neutral.
The centre of gravity remains at constant height.
5.1
(a) first law (b) second law
Solve following problems with the help of above text and
(c) third law (d) All of the above
examples :
6. A cannon after firing recoils due to
1. Swimming is possible on account of
(a) conservation of energy
(a) Newton’s first law of motion
(b) backward thrust of gases produced
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(c) Newton’s third law of motion
(d) Newton’s first law of motion
(d) Newton’s law of gravitation
7. Newton’s second law measures the
2. Inertia is that property of a body by virtue of which the
(a) acceleration (b) force
body is
(c) momentum (d) angular momentum
(a) unable to change by itself the state of rest
8. We can derive Newton’s
(b) unable to change by itself the state of uniform motion
(a) second and third laws from the first law
(c) unable to change by itself the direction of motion
(b) first and second laws from the third law
(d) All of the above
(c) third and first laws from the second law
3. An object will continue moving uniformly when
(d) All the three laws are independent of each other
(a) the resultant force on it is increasing continuously
9. A jet plane moves up in air because
(b) the resultant force is at right angles to its rotation
(a) the gravity does not act on bodies moving with high
(c) the resultant force on it is zero
speeds
(d) the resultant force on it begins to decrease
(b) the thrust of the jet compensates for the force of
4. A man getting down a running bus falls forward because
gravity
(a) of inertia of rest, road is left behind and man reaches
(c) the flow of air around the wings causes an upward
forward
force, which compensates for the force of gravity
(b) of inertia of motion upper part of body continues to
(d) the weight of air whose volume is equal to the volume
be in motion in forward direction while feet come to
of the plane is more than the weight of the plane
rest as soon as they touch the road.
10. When a body is stationary
(c) he leans forward as a matter of habit
(a) there is no force acting on it
(d) of the combined effect of all the three factors stated
(b) the force acting on it is not in contact with it
in (a), (b) and (c).
(c) the combination of forces acting on it balances each
5. A man is at rest in the middle of a pond of perfectly smooth
other
ice. He can get himself to the shore by making use of
(d) the body is in vacuum
Newton’s
ANSWER KEY
1. (c) 2. (d) 3. (c) 4. (b) 5. (c) 6. (c) 7. (b) 8. (c) 9.(b) 10.(c)
COMMON FORCES IN MECHANICS
1. Weight : It is the force with which the earth attracts a body
Case 1 Case 2
and is called force of gravity, For a body of mass m, where 2T
T
acceleration due to gravity is g, the weight 2T
T Massless
W = mg
pulley
T T
2. Tension : The force exerted by the ends of a loaded/stretched
string (or chain) is called tension. The tension has a sense T T
m m
of pull at its ends. 1 2
m g m g
1 2
Laws of Motion 131
Case 3 4. Spring force : If an object is connected by spring and spring
is stretched or compressed by a distance x, then restoring
a
T' T' T T T T1 force on the object F = – kx
m where k is a spring contact on force constant.
T
T – T = ma 5. Frictional force : It is a force which opposes relative motion
T 1
If m = 0, T = T between the surfaces in contact. f = (cid:80)N
1
i.e tension is same
This will be discussed in detail in later section.
The tension in a string remains the same throughout the string if 6. Pseudo force : If a body of mass m is placed in a non-inertial
(a) string is massless, frame having aceleration a(cid:71), then it experiences a Pseudo
(b) pulley is massless or pulley is frictionless force acting in a direction opposite to the direction of a(cid:71).
Case 4 : String having mass
F(cid:71) (cid:32)–ma(cid:71)
pseudo
Negative sign shows that the pseudo force is always directed
in a direction opposite to the direction of the acceleration of
the frame.
y
Let the total mass of the string be M and length be L. Then mass
M a
per unit length is
L F
m pseudo
Let x be the distance of the string from the mass m. Then the mass x
(cid:167)M (cid:183) z
of the shaded portion of string is (cid:168) (cid:117)x(cid:184)
(cid:169) L (cid:185) CONSTRAINT MOTION :
If the string is at rest then the tension T has to balance the wt of When the motion of one body is dependent on the other body, the
shaded portion of string and weight of mass m. relationship of displacements, velocities and accelerations of
the two bodies are called constraint relationships.
(cid:167) M (cid:183)
(cid:63)T(cid:32)(cid:168)m(cid:14) x(cid:184)g Case 1 Pulley string system :
(cid:169) L (cid:185)
(cid:159) as x increases, the tension increases. Thus tension is non- X
uniform in a string having mass. F
3. Normal force : It measures how strongly one body presses
the other body in contact. It acts normal to the surface of x Block
contact.
Step 1 : Find the distance of the two bodies from fixed points.
mg Step 2 : The length of the string remain constant. (We use of
Case 1 N = mg
N this condition)
Therefore X + (X – x) = constant (cid:159) 2X – x = constant
Case 2
dX dx dX dx
a N – mg = ma (cid:159) 2 – (cid:32)0 (cid:159) 2 (cid:32)
dt dt dt dt
m mg (cid:159)(cid:3)(cid:3)N = m(g + a)
(cid:170) dX
(cid:159) 2V (cid:32)v (cid:32)V (cid:32)velocity of pulley
N p B (cid:171)(cid:39) p
(cid:172) dt
dx (cid:186)
Case 3 (cid:32)vB (cid:32)velocity of block(cid:187)
dt (cid:188)
N
(cid:84)
mg sin(cid:84) mg cos (cid:84) Again differentiating we get, 2ap = aB
mg
(cid:170) dVp dv (cid:186)
a (cid:32) and a (cid:32) B
N = mg cos (cid:84) (cid:171)(cid:172) p dt B dt (cid:187)(cid:188)
(cid:84) a = acceleration of pulley, a = acceleration of block
p B
132 Physics
Case 1 : Masses M and M are tied to a string, which goes
Case 2 Here h2(cid:14)x2 (cid:14)y(cid:32)constt. On differentiating w.r.t ‘t’ over a frictionless pu1lley 2
(a) If M > M and they move with acceleration a
2 1
y
h
1 (cid:84) 2 T
F a T
x M
1 a
[Negative sign with dy/dt shows that with increase in time, y
decreases] M g M
1 2
1(cid:117)2x dx dy
(cid:16) (cid:32)0 (cid:159) cos (cid:84) (v – v ) = 0
1 2
2 h2(cid:14)x2 dt dt M g
2
FBD of M , FBD of M
(cid:170) x (cid:186) 1 2
(cid:171)(cid:39)cos(cid:84)(cid:32) (cid:187) T T
(cid:171)(cid:172) h2(cid:14)x2(cid:187)(cid:188)
Cofa wsee d3g We ebdlogcek b alto ct k= s0y satnedm d :o Tttheidn l liinneess aret pt r=e stents the condition M1 a M2 a
M g M g
1 2
c
T(cid:16)M g(cid:32)M a M g(cid:16)T(cid:32)M a
ax 1 1 2 2
where T is the tension in the string. It gives
ay ay M (cid:16)M 2M M
a(cid:32) 2 1g and T (cid:32) 1 2 g
M (cid:14)M M (cid:14)M
(cid:84) 1 2 1 2
(b) If the pulley begins to move with acceleration f,
Ax B ax Ax A downwards
(cid:84)
a(cid:74)(cid:74)(cid:71)(cid:32)M2(cid:16)M1((cid:74)g(cid:74)(cid:71)(cid:16)(cid:74)f(cid:74)(cid:71)) and T(cid:74)(cid:71)(cid:32) 2M1M2 ((cid:74)g(cid:74)(cid:71)(cid:16)(cid:74)f(cid:74)(cid:71))
Ax = acceleration of wedge towards left M1(cid:14)M2 M1(cid:14)M2
a , a = acceleration of block as shown
x y Case 2 : Three masses M , M and M are connected with
1 2 3
a strings as shown in the figure and lie on a frictionless surface.
From (cid:39) ABC , tan(cid:84)(cid:32) y
They are pulled with a force F attached to M .
ax (cid:14)Ax 1
T T T T
Frame of Reference : M3 2 2 M2 1 1 M1 F
Reference frames are co-ordinate systems in which an event is
The forces on M and M are as follows
described. 2 3
There are two types of reference frames T (cid:32) M2(cid:14)M3 F andT (cid:32) M3 F;
(a) Inertial frame of reference: These are frames of reference 1 M (cid:14)M (cid:14)M 2 M (cid:14)M (cid:14)M
1 2 3 1 2 3
in which Newton’s laws hold good. These frames are at rest
with each other or which are moving with uniform speed F
Acceleration of the system is a(cid:32)
with respect to each other. M (cid:14)M (cid:14)M
1 2 3
All reference frames present on surface of Earth are Case 3 : Two blocks of masses M and M are suspended
1 2
supposed to be inertial frame of reference. vertically from a rigid support with the help of strings as shown
(b) Non – inertial frame of reference: Newton’s law do not in the figure. The mass M is pulled down with a force F.
2
hold good in non-inertial reference frame.
All accelerated and rotatory reference frames are non –
T
inertial frame of reference. Earth is a non-intertial frame. 1
T
1
When the observer is in non-inertial reference frame
M g M
a pseudo force is applied on the body under observation. 1 1
T
Free Body Diagram (FBD) : 2
T
Free body diagram of a mass is a separate diagram of that mass. 2
All forces acting on the mass are sketched. A FBD is drawn to M g
2 M
visualise the direct forces acting on a body. 2
F
Laws of Motion 133
The tension between the masses M and M will be (ii) When the mass M moves downwards with
1 2 1
T2 = F + M2g acceleration a.
Tension between the support and the mass M will be
1 Equation of motion for M and M ,
T = F + (M + M )g 1 2
1 1 2 M g sin (cid:84) – T = M a ...(1)
Case 4 : Two masses M and M are attached to a string which 1 1
1 2 T – M g = M a ...(2)
passes over a pulley attached to the edge of a horizontal table. 2 2
The mass M lies on the frictionless surface of the table. Solving eqns. (1) and (2) we get,
1
T
(cid:170)M sin(cid:84)(cid:16)M (cid:186) (cid:170) M M (cid:186) g
M1 a(cid:32)(cid:171) 1 2(cid:187)g;T (cid:32)(cid:171) 2 1 (cid:187)
(cid:172) M1(cid:14)M2 (cid:188) (cid:172)M1(cid:14)M2(cid:188)(1(cid:14)sin(cid:84))
a
(a) If (M /M = sin(cid:84)) then the system does not accelerate.
T 2 1
(b) Changing position of masses, does not affect the
M tension. Also, the acceleration of the system remains
2
unchanged.
M2g
(c) If M = M = M (say), then
1 2
Let the tension in the string be T and the acceleration of the
2 2
system be a. Then (cid:167) (cid:84) (cid:84)(cid:183) (cid:167)g(cid:183) (cid:167) (cid:84) (cid:84)(cid:183) (cid:167)Mg(cid:183)
T = M1a ...(1) a(cid:32)(cid:168)(cid:169)cos2(cid:16)sin2(cid:184)(cid:185) (cid:168)(cid:169)2(cid:184)(cid:185); T (cid:32)(cid:168)(cid:169)cos2(cid:14)sin2(cid:184)(cid:185) (cid:168)(cid:169) 2 (cid:184)(cid:185)
M g – T = M a ...(2)
2 2
Adding eqns. (1) and (2), we get Case 6 : Two masses M1 and M2 are attached to the ends of a
string over a pulley attached to the top of a double inclined
(cid:170) M (cid:186) (cid:170) M M (cid:186)
a(cid:32)(cid:171) 2 (cid:187)g and T (cid:32)(cid:171) 1 2 (cid:187)g plane of angle of inclination (cid:68) and (cid:69).
(cid:172)M1(cid:14)M2(cid:188) (cid:172)M1(cid:14)M2(cid:188) Let M move downwards with acceleration a and the tension in
2
Case 5 : Two masses M and M are attached to the ends of a the string be T then
1 2
string, which passes over a frictionless pulley at the top of the
inclined plane of inclination (cid:84). Let the tension in the string be T.
M M
1 2
N
M (cid:68) (cid:69)
1
M
M1g sin(cid:84) M1g(cid:84) M1g cos(cid:84) 2 FBD of M1 a T
M g M
2 1
(cid:84)
(i) FWrohmen tthhee FmBaDss Mof 1M m1o avnesd uMpw2,ards with acceleration a. M 1gsin(cid:68) M(cid:68)1gM1gcos(cid:68)
T – M g sin (cid:84) = M a ...(1)
1 1 Equation of motion for M
1
M g – T = M a ...(2)
2 2 T – M g sin (cid:68) = M a
1 1
Solving eqns. (1) and (2) we get,
or T = M g sin (cid:68) + M a ...(1)
1 1
a(cid:32)(cid:170)(cid:171)M2(cid:16)M1sin(cid:84)(cid:186)(cid:187)g FBD of M2 T a
(cid:172) M1(cid:14)M2 (cid:188) M
2
FBD of mass M
1
Mgcos(cid:69) (cid:69) M
R=N T y x 2 M2g 2gsin(cid:69)
Equation of motion for M
2
M g sin(cid:69) – T = M a
2 2
or T = M g sin (cid:69) – M a ...(2)
Mg cos (cid:84) 2 2
1
Mg sin (cid:84) Mg Using eqn. (1) and (2) we get,
1 1
M g sin (cid:68) + M a = M g sin (cid:69) – M a
FBD of M 1 1 2 2
2 Solving we get,
(cid:170) M M (cid:186) g
T(cid:32)(cid:171) 2 1 (cid:187) (cid:11)M sin(cid:69)(cid:16)M sin(cid:68)(cid:12)g M M g
(cid:172)M1(cid:14)M2(cid:188)(1+sin(cid:84)) T a a(cid:32) 2 1 and T (cid:32) 1 2 [sin(cid:69)(cid:14)sin(cid:68)]
M (cid:14)M M (cid:14)M
Mg 1 2 1 2
2
134 Physics
Case 7 : A person/monkey climbing a rope (d) If angles of inclination are (cid:84) and (cid:84) for two inclined planes
1 2
T ½
t (cid:167)sin(cid:84) (cid:183)
a Keeping the length constant then 1 (cid:32)(cid:168) 2(cid:184)
t2 (cid:169)sin(cid:84)1 (cid:185)
Case 9 : Weight of a man in a lift :
a
(i) When lift is accelerated upward : In this case the man also
Mg moves in upward direction with an acceleration a(cid:71).
(a) A person of mass M climbs up a rope with acceleration a.
The tension in the rope will be M(g+a).
T – Mg = Ma (cid:159) T = M(g + a)
(b) If the person climbs down along the rope with acceleration a a
a, the tension in the rope will be M(g–a).
mg
T
N
Then from Newton’ second law
N – mg = ma or N = m(g + a)
a
a
or W = m(g + a) (cid:32)W (1(cid:14)a/g) (as W = mg)
app o
Mg
Where W is apparent weight of the man in the lift, W is
app o
Mg – T = Ma (cid:159) T = M(g – a) the real weight, N is the reaction of lift on the man. It is clear
(c) When the person climbs up or down with uniform speed, that N = W
app
tension in the string will be Mg. When the lift moves upward and if we measure the weight
Case 8 : A body starting from rest moves along a smooth inclined of the man by any means (such as spring balance) then we
plane of length l, height h and having angle of inclination (cid:84). observe more weight (i.e., W ) than the real weight (W )
app o
W >W
app o
(ii) When lift is accelerated downward : In this case from
l Newton’s second law
h FBD of body
(cid:84)
a
N=R
mg
N
(cid:84)
mg – N = ma
mg sin(cid:84) mg cos(cid:84) or N = m(g – a) = W (1– a/g)
mg o
or W' = W (1– a/g) (cid:94) W (cid:32)mg(cid:96)
(where N=R is normal reaction applied by plane on the body app o (cid:39) o
If we measure the weight of man by spring balance, we
of mass m)
For downward motion, along the inclined plane, observe deficiency because Wapp< Wo.
mgsin(cid:84)(cid:32)ma(cid:159)a(cid:32)gsin(cid:84)
(iii) When lift is at rest or moving with constant velocity : From
By work-energy theorem loss in P.E. = gain in K.E.
Newton’s second law N –mg = 0 or N = mg
1 In this case spring balance gives the true weight of the man.
(cid:159)mgh (cid:32) mv2 (cid:159)v(cid:32) 2gh
2 Case 10 : Three masses M , M and M are placed on a smooth
1 2 3
surface in contact with each other as shown in the figure.
Also, from the figure, h = sin (cid:84). (cid:63) v(cid:32) 2gh (cid:32) 2g sin(cid:84)
(cid:65) (cid:65) A force F pushes them as shown in the figure and the three
(a) Acceleration down the plane is g sin (cid:84). masses move with acceleration a,
(b) Its velocity at the bottom of the inclined plane will be
2gh (cid:32) 2g sin(cid:84)
(cid:65) M
3
(c) Time taken to reach the bottom will be M
2 M
F 1
t(cid:32)(cid:168)(cid:167)(cid:169)gs2i(cid:65)n(cid:84)(cid:184)(cid:183)(cid:185)1/2 (cid:32)(cid:168)(cid:168)(cid:167)(cid:169)gs2inh2(cid:84)(cid:184)(cid:184)(cid:183)(cid:185)1/2 (cid:32) (cid:167)1g (cid:183)1/2 (cid:32)si1n(cid:84) 2gh F2 F21 F1 F
sin(cid:84)(cid:168) (cid:184)
(cid:169)2h(cid:185) a
Laws of Motion 135
M
1 (cid:159) F – F = m a ...(i) (cid:32) mgR(cid:170)1(cid:16)cos (cid:65)(cid:186) ;
F F 1 1 (cid:171) (cid:187)
1 (cid:65) (cid:172) R(cid:188)
M
F2 2 F1 (cid:159) F1 – F2 = m2a ...(ii) (cid:63) Acceleration, a(cid:32) F (cid:32)gR(cid:167)(cid:168)1(cid:16)cos (cid:65) (cid:183)(cid:184) .
M3 m (cid:65) (cid:169) R(cid:185)
(cid:159)(cid:3)F = M a ...(iii)
F 2 3 Example 2.
2
Having gone through a plank of thickness h, a bullet
F
Adding eqns. (i), (ii) and (iii) we get, a(cid:32) changed its velocity from u to v. Find the time of motion of
M (cid:14)M (cid:14)M
1 2 3 the bullet in the plank, assuming the resistance force to be
M F (M (cid:14)M )F proportional to the square of the velocity.
(cid:159)(cid:3)F2 (cid:32) 3 and F1(cid:32) 2 3 Solution :
M (cid:14)M (cid:14)M M (cid:14)M (cid:14)M
1 2 3 1 2 3
Given force F= (cid:16)kv2, where k is a constant. Negative sign
Keep in Memory shows that the force is retarding one. Now, force = rate of
change of momentum = m dv/d t ;
1. When a man jumps with load on his head, the apparent
weight of the load and the man is zero. mdv (cid:32)(cid:16)kv2 or mdv/v2 (cid:32)(cid:16)kdt ;
2. (i) If a person sitting in a train moving with uniform dt
velocity throws a coin vertically up, then coin will fall
Integrating it within the conditions of motion i.e. as time
back in his hand.
changes from o to t, the velocity changes from u to v, we
(ii) If the train is uniformly accelerated, the coin will fall
have
behind him.
(iii) If the train is retarded uniformly, then the coin will fall v t v
dv (cid:167)1(cid:183)
in front of him. m(cid:179) (cid:32)(cid:16)k(cid:179)dt ; or (cid:16)m(cid:168) (cid:184) (cid:32)(cid:16)kt
v2 (cid:169)v(cid:185)
Example 1. u 0 u
A chain of length is placed on a smooth spherical surface
(cid:65)(cid:1)
of radius R with one of its ends fixed at the top of the m(cid:167)u(cid:16)v(cid:183)
or t (cid:32) (cid:168) (cid:184) ....(i)
sphere. What will be the acceleration a of each element of k (cid:169) uv (cid:185)
the chain when its upper end is released? It is assumed
dv dv ds
(cid:167) R(cid:183) Also, F(cid:32)m (cid:32)m (cid:32)(cid:16)kv2
that the length of chain (cid:65)(cid:31)(cid:168)(cid:652) (cid:184). dt ds dt
(cid:169) 2(cid:185)
Solution : dv k (cid:167) ds (cid:183)
Let m be the mass of the chain of length . Consider an or (cid:32)(cid:16) ds. ; (cid:168)(cid:39) (cid:32)v(cid:184) ; Integrating it,
(cid:65) v m (cid:169) dt (cid:185)
element of length d of the chain at an angle (cid:84) with vertical,
(cid:65)
v h
dv k k
we get (cid:179) (cid:32)(cid:16) (cid:179)ds or(cid:11)log v(cid:12)v (cid:32)(cid:16) (cid:11)s(cid:12)h
v m e u m o
u 0
dl
d(cid:84) (cid:16)k kh
(cid:84) or log v(cid:16)log u(cid:32) (h(cid:16)0)(cid:32)(cid:16)
e e m m
R
m
or k(cid:32) log (u/v)
h e
From figure, d = R d (cid:84) ;
(cid:65) Putting this value in eqn. (i), we get
Mass of the element,
m(u(cid:16)v)/uv h(u(cid:16)v)
m m t(cid:32) (cid:32)
dm = d ; or dm = .Rd(cid:84)
(cid:65) (m/h) log (u/v) uv log (u/v)
(cid:65) (cid:65) e e
Force responsible for acceleration, dF = (dm)g sin(cid:84) ; Example 3.
A wire of mass 9.8 × 10-3 kg per metre passes over a
(cid:167)m (cid:183) mgR
dF = (cid:168) Rd(cid:84)(cid:184)(gsin(cid:84))(cid:32) sin(cid:84)d(cid:84) frictionless pulley fixed on the top of an inclined
(cid:169) (cid:65) (cid:185) (cid:65) frictionless plane which makes an angle of 30º with the
Net force on the chain can be obtained by integrating the horizontal. Masses M and M are tied at the two ends of
1 2
above relation between 0 to (cid:68), we have the wire. The mass M rests on the plane and mass M
1 2
(cid:68)mgR mgR (cid:68) mgR hangs freely vertically downwards. The whole system is in
F(cid:32)(cid:179) sin(cid:84)d(cid:84)(cid:32) ((cid:16)cos(cid:84)) (cid:32) [1(cid:16)cos(cid:68)] equilibrium. Now a transverse wave propagates along the
(cid:65) (cid:65) 0 (cid:65) wire with a velocity of 100 ms–1. Find M and M .
0 1 2
136 Physics
Solution :
1 1
Resolving M g into rectangular components, we have M g h(cid:32)0(cid:14) g(t/2)2 (cid:32) gt2/4 ...(ii)
1 1 2 2
sin 30º acting along the plane downwards, and M g cos30º
1 Dividing eqn. (ii) by (i), we get
acting perpendicular to the plane downwards. The situation
has been shown in fig. h 1
(cid:32) [ cos(cid:84)(cid:32)h/ ]
(cid:39) (cid:65)
4cos(cid:84)
(cid:65)
R 1 1 1
or cos(cid:84)(cid:32) ; or cos2(cid:84)(cid:32) ; or cos(cid:84)(cid:32)
T 4cos(cid:84) 4 2
F
g sin (cid:84) or (cid:84) = 60º
M (cid:84)
1 Mg cos(cid:84) Example 5.
Mg 1 Mg
(cid:84) 1 2 A large mass M and a small mass
m hang at the two ends of a string
Let T be the tension in the wire and R be the reaction of
that passes through a smooth
plane on the mass M . Since the system is in equilibrium,
1 tube as shown in fig. The mass m
therefore,
moves around a circular path in
T = M g sin30º ...(i) (cid:84)
1 a horizontal plane. The length of l
and R = M g cos30º ...(ii)
1 the string from mass m to the top
T = M g ...(iii) r m
2 of the tube is l, and (cid:84) is the angle
From eqn. (i) and (iii) we have
the string makes with the
T = M g sin30º = M g ...(iv)
1 2 vertical. What should be the
T frequency ((cid:81)) of rotation of mass
Velocity of transverse wave, v(cid:32) , m so that mass T
m
M remains stationary? M
where m is the mass per unit length of the wire.
Solution :
(cid:63) v2 (cid:32)T/m, or T = v2m = (100)2 × (9.8 × 10-3) = 98N Tension in the string T = Mg.
From eqn . (iii), M = T/g = 98/9.8 = 10kg. Centripetal force on the body = mr(cid:90)2 =mr ( 2(cid:83)(cid:81))2. This is
2
From eqn . (iv), M = 2M = 2 × 10 = 20kg. provided by the component of tension acting horizontally
1 2 i.e. T sin(cid:84) ( = Mg sin(cid:84)).
Example 4.
A block slides down a smooth inclined plane to the ground 1 Mg
when released at the top, in time t second. Another block (cid:63) mr ( 2(cid:83)(cid:81))2 = Mg sin(cid:84) = Mgr/l. or (cid:81)(cid:32)
2(cid:83) ml
is dropped vertically from the same point, in the absence
Example 6.
of the inclined plane and reaches the ground in t/2 second.
A string of negligible mass going over a clamped pulley of
Then find the angle of inclination of the plane with the
mass m supports a block of mass M as shown in fig. The
vertical.
force on the pulley by the clamp is given by
Solution :
If (cid:84) is the angle which the inclined plane makes with the (a) 2Mg
vertical direction, then the acceleration of the block sliding m
down the plane of length will be g cos(cid:84). (b) 2 mg
(cid:65)
A (c) [ (M(cid:14)m)2(cid:14)m2]g
(cid:84)
(d) [ (M(cid:14)m)2(cid:14)M2]g M
(cid:65) h
Solution : (c)
Force on the pulley by the clamp = resultant of
T = (M + m)g and mg acting along horizontal and vertical
C B
respectively
Using the formula, s(cid:32)ut(cid:14)1at2, we have s = , u = 0, t = (cid:63) F(cid:32) [(M(cid:14)m)g]2(cid:14)(mg)2 (cid:32)[ (M(cid:14)m)2(cid:14)m2]g
(cid:65)
2
Example 7.
t and a = g cos (cid:84).
The masses of 10 kg and 20 kg respectively are connected
1 1 by a massless spring in fig. A force of 200 newton acts on the
so (cid:65)(cid:32)0(cid:117)t(cid:14) gcos(cid:84)t2 (cid:32) (gcos(cid:84))t2 ...(i) 20 kg mass. At the instant shown, the 10 kg mass has
2 2
acceleration 12 m/sec2. What is the acceleration of 20 kg
Taking vertical downward motion of the block, we get
mass?
Laws of Motion 137
20 kg
10 kg m
200 newton
M'
Solution : (cid:84)
Force on 10 kg mass = 10 × 12 = 120 N
M
The mass of 10 kg will pull the mass of 20 kg in the backward
direction with a force of 120 N.
Solution :
(cid:63) Net force on mass 20 kg = 200 – 120 = 80 N
Since m does not slip on M' (relative velocity of m w.r.t. M'
force 80N is zero)
Its acceleration a (cid:32) (cid:32) (cid:32)4m/s2
mass 20kg (cid:63)(cid:3)M', m will move with same acceleration as that of M.
Since surfaces are smooth
Example 8.
Two masses each equal to m are lying on X-axis at (–a, 0) (cid:63) frictional force is zero
and (+ a, 0) respectively as shown in fig. They are connected Net force = Mg = (M + M' + m) a
by a light string. A force F is applied at the origin and
Mg
along the Y-axis. As a result, the masses move towards each (cid:63)a (cid:32) ....(1)
other. What is the acceleration of each mass? Assume the M(cid:14)M(cid:99)(cid:14)m
instantaneous position of the masses as (– x, 0) and (x, 0) Now let us see m, w.r.t. M'
respectively
F
N
ma
mg
(cid:84)
(–a, 0) (a, 0)
–X X Downward acceleration of m on slope = 0
m O m
Solution : (cid:63) N – ma sin (cid:84) + mg cos (cid:84) = 0 ....(2)
(net (cid:65) force = 0)
F
and mg sin (cid:84) – ma cos (cid:84) = 0 ....(3)
A [ net force along slope = 0]
(cid:39)
From eqn. (3) g sin (cid:84) = a cos (cid:84)(cid:3)(cid:3)or a = g tan (cid:84) ....(4)
T (cid:84) T
From eqn. (4) and (1),
B C
(–x, 0) O (x, 0) M
we have tan(cid:84)(cid:32) (cid:159) M cot (cid:84) = M + M' + m
M(cid:14)M(cid:99)(cid:14)m
From figure F = 2 T cos (cid:84) or T= F/(2 cos (cid:84))
The force responsible for motion of masses on X-axis is T M(cid:99)(cid:14)m
(cid:159)M(cid:32)
sin (cid:84) cot(cid:84)(cid:16)1
Example 10.
F
(cid:63)ma(cid:32)T sin(cid:84)(cid:32) (cid:117)sin(cid:84)
2cos(cid:84) Find the acceleration of block A and B. Assume pulley is
massless.
F F OB F x
(cid:32) tan(cid:84)(cid:32) (cid:117) (cid:32) (cid:117)
2 2 OA 2 (a2(cid:16)x2) 2kg
A
F x
so, a(cid:32) (cid:117)
2m (a2(cid:16)x2)
Example 9.
Find the mass M of the hanging block in figure which will
prevent smaller block from slipping over the triangular B 5kg
block. All surfaces are frictionless and the string and the
pulley are light.