Hints and Solution LOM
Short Description
LAWS OF MOTION...
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Laws of Motion Solutions to Subjective Assignments LEVEL – I 1.
We find that the each of the blocks A & B experience following forces. Forces acting on A : (a) Earth's gravitational field force = weight of the block A = M1g (b) Normal contact force exerted by the horizontal surface = N1. (c) Contact force exerted by the block B = N (d) The external horizontal force = F. Forces acting on B : (a) Earth's gravitational field force = weight of the block B = M2g (b) Normal contact force exerted by the horizontal surface = N2. (c) Contact force exerted by the block A = N Both the blocks move together. Hence both will have same acceleration, let us chose X-axis shown. There is no Y-axial acceleration as the blocks are not coming off the surface. The F.B.D. diagram of ‘A’ as a system is shown in figure (i) . The F.B.D. of ‘B’ as a system shown in figure (ii). N1
N2
F
N
N
a
a
M1g
M2g
Fig. (i)
From figure (i), For A,
N1 – M1g = 0 F – N = M1a
Fig. (ii)
…(1) …(2)
From figure (ii) N2 – M2g = 0 …(3) For B, N = M2a …(4) Thus solving (1), (2), (3) and (4) We obtain F M2F and N a= M1 M2 M1 M2 2.
Along vertical direction, T cos = mg Along horizontal direction T sin = ma Using (1) and (2) tan = a/g
T cos
…(1)
a
…(2)
T sin mg
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a = g tan = 9.8 tan 30o =
9. 8 3
m / s2
Therefore, acceleration of the trolley is a= 3.
9 .8 3
m / s2 =5.7 m/s2.
The F.B.D. of m1 gives, N1 = m1g Cos T = m1g Sin The F.B.D. of m2, gives, N2 = m2g Cis T = m2g Sin
…(1) …(2) …(3) …(4)
N1
T T
N2
m1
m2 m1 g cos
m1 g sin
m2 g cos
m2 g sin
Taking the ratio of (2) and (4), m 1 Sin Sin 30o 1 2 1 = . o = 2 m2 Sin Sin 60 3 3
4.
m1 1 . m2 3
Let the block be displaced to the left by x the forces acting on the block will be kx to the right by both the springs thus F = ma
5.
or
a=
F 2kx m m
fmax .2 1 9.8 N = 1.96 N (a) When F = 1N then, F < fmax and f = F f = 1N (b) When F = 1.96 N, f = fmax = 1.96 N
F = fmax
(c) When F = 2.5 N F > fmax f = fk = kN
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= kmg = 0.15 1 9.8 = 1.47 N.
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6.
F = (m1 + m2)a F 1 1 a= = m/s2 m1 m2 5 2 7 if the string breaks, m2 will move with constant velocity attained by it while m1 will move with a acceleration 1 a = m/s2 5
7.
(a) Angle of repose = tan1() = tan1(0.3) = 16.7 (b) Frictional force = smgcos(/2)
8.
16.7
ffr = 0.3 cos
2
mg = 0.145 mg
F
(a) Let the normal contact force be N F sin + N = mg N = mg – F sin = 20 10 – 80 sin 300 = 160 N
m N
fk
(b) F cos = 80 cos 30 = 40 3 fk = k N = (0.3 ) (160) = 48 N 0
(c) a = (Fcos - fk)/m =
mg
40 3 48 = 1.06 m/s2. ( 20 )
(d) F' cos = fk F' = fk sec = 48 (2 / 3) = (96/3) = 55.42 N N
kN
9. Coefficient of friction, k = 0.14 N = 0.14 N – mg cos = 0 mg sin - kN = ma or, mg (sin - k cos) or, a = g (sin - k cos ) The distance travelled by the body is given by 1 2 at = 2 cos or,
t=
A
mg
2 2 a cos g(sin k cos
for t to be a minimum, dt 0 tan 2 = - 1/k d The minimum value of t is
g 1 k2 k 2
1
2
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10.
Taking the differential element at any angular position , we obtain The net tangential force = (dm)g sin
=
M .( ds ) g sin l 0
R d
dm
(dm)g
M l .(Rd) g sin 0
MRg sin d l 0 MRg cos 0 , = l where R = l = l/R . Tangential acceleration = Rg (1 cos ) . R
=
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LEVEL – II 1.
The forces acting on the particle are mg (vertically downward) ma0 (pseudo force vertically downward) N normal reaction, normal to the inclined plane) Resolving the forces along the incline, we have F = m (a0 + g) sin acceleration down the plane g 3g g sin 300 = 4 2
a = (a0 + g) sin =
Distance traversed along the incline 1 3g 2 t L sec = at 2 = 2 8 t=
8L sec 30 3g
16L 3 3g
2.
F Motion of the body up the plane F = mg sin + mg cos m mg sin = mg[ sin + cos ] mg cos Motion of the body down the plane F'= mg cos - mg sin m F' mg cos = mg |( cos - sin ) | mg sin Put = 0.2, = 300, & m = 2 kg Note : If F' is negative alter its assumed direction.
3.
Acceleration of the man relative to the elevator = 2 m/s2 (upwards) Acceleration of the elevator relative to earth = 3 m/s2 (downwards) Hence, acceleration of the mass relative to earth = 3 – 2 = 1 m/s2 (downwards) Consider free body diagram of man : If T = tension in the spring ; mg – T = ma 50 10 – T = 50 1 T = 450 N. 450 Hence, extension in the spring = = 0.225 m. 2000
4.
mv
dv mg d cos = ds = ds 3 a v g vdv = cos d 3 a0 0
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v2 g sin 2 3a 2g v= sin . 3a
5.
Acceleration of a block sliding down an inclined plane is given by a = g ( - cos ) for 2 kg block 1 < 2 , hence a1 > a2 Hence the blocks move separately and with unequal accelerations. Acceleration of 2 kg block is a = g(sin 30 – 0.2 cos 30) = 10 (0.5 – 0.1732) = 3.27 m/s2 In case (b), when 1 = 0.3 and 2 = 0.2, the 2 kg block has less acceleration than the 3 kg block. Thus the blocks move together. When we draw the FBD and resolved the forces along the plane, we get F = (m1 + m2) g sin - (m1 + m22) g cos a=
(m11 m2 2 ) F g cos = g sin m1 m2 m1 m2
= 10 sin 30 -
6.
7.
( 2 0 .3 4 0 .2 ) 10 cos 30 = 2.97 m/s2 6
Initially mg – F = ma . . . (i) After releasing a mass (m) F – (m- m)g = (m - m) a . . .(ii) Where F = upward thrust of air By solving (i) and (ii) & putting mg = w, obtain the answer. Let there is no relative motion between the blocks B and C Hence T = (mB + mC)a (1) And mAg - T = mAa (2) From (1) and (2), we get
mA g 30 5 m / s2 a= m A mB mC 18 3 Net force on the block C is F = mCa = 10 (5/3) N = 16.6 N
F
F a
mg
a (m-m)g
F.B.D. of the blocks NB T
fr mB g
T
NC fr
mAg
NB+mCg
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Since force acting on the block C is only frictional force. If maximum value of frictional force acting on C is fmax , then f(max) = mBg = 0.4 5 10 = 20 N F fmax Hence there is no relative motion between the block B and C. Therefore distance moved by C is 2m only. 8.
Since string is massless and friction is absent hence tension in the string is same, every where. (a) Let acceleration of the pulley be aP. For aP to be F non zero. F 2T. . . (1) also T m1g T 2g . . . (2) From (1) and (2), we get F 2 (2g) T T F 40 N Therefore when F = 35 N aP = 0 and hence a1 = a2 = 0 (b) as mass of the pulley is negligible F - 2T = 0 T = F/2 T = 35 N To lift m2, T m2g T 50 N Therefore block m2 will not move F.B.D. of m1
F
yP
m1 m2
y1
T = 35N
m1g = 20 N
T - m1g = m1a1 15 = 2a1 15 m / s2 a1 = 2 Constraint equation, yP + yP - y1 = constant 2yP - y1 = c
aP =
2
d2 y p dt 2
d2 y1 2 0 dt
a1 15 m / s2 2 4
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(c) When F = 140 N T = 70 N F
T m1
a1 m1
m1g
T - m1g = m1a1 (1) 70 N - 20 N = 2 a1 a1 = 25 m/s2 T - m2g = m2a2 70N - 50 N = 5a2 a2 = 4 m/s2 Constraint equation yP - y2 + yP - y1 = c 2yP - y1 - y2 = c
9.
yP
m2 y1 y2
(1)
d2 y P d2 y 1 d2 y 2 2 2 2 0 dt dt dt
aP =
a1 a2 29 m / s2 2 2
Equation of motion for m: Fx = ma N = ma …(1) Fy = 0 Since m is at rest relation to M at the verge of slipping. f – mg = 0 Equation of motion for M: F – N – N = Ma mg mg , Where N = ,a=
…(4) a
g
N a N mg N a
N
F
and since the block
M does not have vertically motion. N = Mg + f Putting values of N, N and a in (4), we obtain F – (Mg + f) – N = Ma F - (Mg + mg) – ma = Ma F - g (M + m) = (M + m)a F g(M m) a= Mm f fmax mg N mg ma
N N
Mg
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F g(M m) Mm (M + m) g F - 2 g (M + m) 2g(M + m) - F + (M + m) g 0 22 - 5 + 2 0 For critical, 22 - 5 + 2 = 0 ( - 2) (2 - 1) = 0 1 = 2 or = 2 Practically does not become 2. = 0.5. g .
10.
Let x1 distance of the pulley (wedge) from the wall. x2 distance of the bar from the pulley. Therefore x1 + x2 = l
d2 x1 dt 2 d2 x1 dt 2
x2
y
=
d2 x 2 dt 2 Acceleration of the bar a cos
dt 2
d2 x 2 dt 2
a sin
a
= a (say).
N
Net acceleration of wedge along the horizontal surface = a …(1) Net acceleration of bar along the horizontal surface figure (i) = a – comp. of acceleration of bar w.r.t. wedge = a – a cos …(2) Vertically downward = a sin …(3) Forces on wedge and bar are shown separately in figure (ii) and fig. (iii) respectively (iv) and (v). From the F.B.D. of the wedge Along the chosen X-direction T + N1 sin – T cos = Ma …(4) along Y-direction For bar
a
wedge =
x
x1
a
acceleration of wedge
d2 x 2 acceleration of the bar, w.r.t. dt 2 d2 x1
a
T N1
T a
Mg
N1
a-acos a sin
mg
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N1 cos + T sin + Mg – N = 0 …(5) From the F.B.D. of the bar along X-direction cos – N1 sin = m (a – a cos ) …(6) For Wedge: along Y-direction mg – N1 cos – T sin = m a sin …(7) From equations (4), (5), (6) and (7) eliminating other quantities we can find a.The problem can be solved by choosing X-Y axis for the system along and normal to the surface of wedge. Just do by yourself.
N1 cos
T sin T cos
N1 sin
a-a cos mg a sin N
T N1 sin
T cos
N1 cos T sin Mg
a
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Solutions to Objective Assignments LEVEL – I 2.
F F 3 3m
T2 = ma = m
a
F – T1 = ma T1 = F - ma F 2F =F3 3 T2 F/3 1 . T1 2F / 3 2 3.
m
T2
T2
The contact force F' accelerates the body of mass m1 = 2 kg with an acceleration F 3 1m / s a= m1 m2 1 2 F = m1a = 1 1 = 1N. Fimpact
Y m
T1
Z
m
T1
F
m2 m1 1 kg
F
2 kg F'
m1
v( dm / dt ) (5)(1) = = 2.5m/s2. m 2
4.
a=
5.
In uniform circular motion Fv where F is the centripetal force 1 KE = mv 2 cons tan t 2
6.
The net force F = M1g sin - M2 g sin = (M1 sin - M2 sin )g When M1 = M2 and 1 = 2 , the net force is zero. The acceleration of the system = 0
7.
Using the expression obtained in question no. 1 We obtain = 1- (1/n2), putting n = 2 we obtain = 0.75
8.
The acceleration of m along the plane is g sin because the inclined plane is smooth The component of the tension T parallel to the inclined plane is zero. The only driving force is component of gravitational force mg parallel to the plane, that is equal to mg sin . Therefore, for the equilibrium of the body, mg cos = T.
m
=
a
a X
T
m
mg sin
mg cos mg
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9.
Since block has been pushed and is brought to motion hence kinetic friction will act mgsin fr = ma fr = 2mgcos gsin 2g cos = a a = g(sin 2cos)
10.
a=
mass v ( v ) = m
= 11.
Fimpact
=
v(dm / dt ) m
v 2 (numerically) m
Following the previous procedure F – T2 = m3a F F – T2 = m3 m1 m2 m3 T2 =
(m1 m2 )F m1 m2 m3
T2 =
(1 8) 36 9 N (1 8 27)
12.
Change in linear momentum = impulse = Area of F/t graph = (1/2) 8 1 = 4 kg-m/s. (or N - s.)
13.
Equation of laws of motion for M : T = Ma . . . (i) Equation of laws of motion for M: F – T = ma . . . .(ii) Eliminating (a) in (i) & (ii) obtain MF T= Mm
14.
a m1
T1
a
a T2
m2
T1
T2
m3 F
a m
M
a M
F
T
a T
m
F
The tension = weight of the chain of length x = mg lx M . (l x )g = Mg = l l
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15.
For the equilibrium of the body m. T cos = mg For horizontal acceleration of m: T sin = ma
T cos
T sin
T sin ma T cos mg
a
= tan1 (a/g). 16.
17.
For equilibrium of the body (block) f – mg = 0 f = mg = (0.1) (9.8) = 0.98 N The maximum frictional force = fmax = N = (0.5)(5) = 2.5 N Since fmax > mg, the block remains in equilibrium and the necessary frictional force developed is equal to mg. The maximum force between ground and B = (mA + mB) g fmax = (0.5) (2 + 8) (10) = 50 N
mg f
N N mg
A
Since the dragging force is F = 25 N Therefore the system does not move that means the accelerations of A and B will be zero. The frictional force between A and B = 0. 18.
Fav =
T
F
B
P t
10
2mv 100 100 2 5 = = = 1000 t 1000
0.01
= 10 N. 19.
20.
T , M where F – T = Ma T = F – Ma F Ma F a. a' = M M
a' =
a' A
T
a B
F
T
Using the expression obtained in question no. 1. 1
1 cot
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cot = 1 -
1 n2
1 = 1 2 tanθ . n LEVEL – II
F M
1.
Acceleration of M, a
2.
1F 2 .t 2M 2Ml . t F Mg – T = Ma …(i) T = ma …(ii) Solving (i) and (ii) Mg a (M m) l
N a M g
FBD of man Mg – N = Ma Mmg N (M m) 3.
M' g T M'a T = Ma
…(i)
…(ii)
M' g a(M M') M' g a (M M') ma sin mg cos a g cot R T
M
M g
M' g (M M') cot M cot M' M' gcot
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M'
Mcot (1 cot )
T=Ma = M. g cot
T
Mg . tan
m a sin
m a m g c o s
4. 5.
m g s in + m acos
If the tendency of relative motion along the common tangent does not exist, then component of contact force along common tangent will be zero. For the two values of F, i.e., F = 150 N and F = 120 N; the tensions in the string are
T
F = 75 N and 60 N. In the first case, the accelerations of the two masses are equal 2
and opposite, while in the second the accelerations are zero. 6.
T1 = 2 mg F = 9 mg (a), (b).
…(i) …(ii) 2m g
T
2m g F
7.
kx = 120 N,
a
9.
10.
2 m g 2 m g m g
200 120 4 m / s2 . 20
120
20kg
200
Since there is relative acceleration between the frames. S1 may be at rest S2 is accelerating S1 may be acceleration and S2 is at rest S1 and S2 both are accelerating with difference acceleration S1 and S2 both cannot be at rest.
r N1 m | a1 | sin mgcos
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N 1 + m | a 1 |s i n m | a1 | | a1 |
m g s i n + m | a 1 |c o s
m g c o s
r N1 m g cos | a1 | sin r N1 sin M | a1 | FBD of m
N 1+ m | a 1 |s in
m g sin
m gcos
N 2 mg N1 cos r N 2 mg m(g cos | a1 | sin ) cos N 2 (M m)g . N
2
N 1s in N 1c o s
m g
COMPREHENSION
1.
mv 2 S mg , since friction is static. r
2. Normal reaction decreases, and so the limiting friction is lower. 3. Net force = f – D = 0. 4. to 6. F.B.D. of man N T
a
M g
N + T – Mg = Ma With N = Mg T = Ma …(i)
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F.B.D. of box :
T a
N
( m + m ’) g
T – N – mg – mg = (m + m)a …(ii) Where M = 60 kg, m = 20 kg, m = 30 kg Solving equation (i) and (ii), we get a = 110 ms–1 T = 6600 N. MATCH THE FOLLOWING 1.
When body is at rest then friction is static and if body is moving then friction on it is kinetic.
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