8.
Solved Problems
8.1
Subjective
Problem 1: A particle of mass m, originally at rest, is subjected to a force whose direction is constant but whose magnitude varies with the time according to the relation F = F0 2 T T t 1 ,
where F0 and T are constants. The force acts only for the time interval 2T. Prove that the speed v of the particle after a time 2T has elapsed is equal to 4F0T/3 m. Solution : F = F0 2 1 T t 1 v = 1 dt T t 1 m F 2T 0 2 0
=
1 dt T t t m F 2T 0 2 T 2 0 0 = T 2 0 3 0 1 T t 3 T T 2 m F =
1(1)
3 T T 2 m F0 = m 3 T F 4 0 .Problem 2: A mass of 10 kg lies on a smooth table at a distance of 7 m from the edge and is connected by a taut string passing over the end with a mass of 4 kg hanging freely. How long does it take the 10 kg mass to reach the edge of the table? Solution : m1 = 4m2 m1g – 2T = m1 2 a T – m2g = m2a (m1 – 2m2)g = (m1/2 + 2m2)a or, a =
g
m
2
2
m
m
2
m
2 1 2 1
= 2 g = 5 m/s2 The mass m1 reaches the ground with a velocity, v, such that,20 cm m1 m2 mg T T T v2 = 2 (a/2) s = 2 2 5 0.2 = 1 or, v = 1 m/s
The velocity of m2 when m1 reaches the ground is 2v = 2m/s.
The maximum height reached by m2 is
h = 0.4 m + m
10 2
22
Problem 3: In the given figure all the surfaces are smooth. Find the time taken by the block to reach from the free end to the pulley attached to the plank. Distance between free end and pulley is .
F m M
Solution: Equation of motion for M :
2F = M a2 …(1) For m, F = m a1 …(2) Using (1) and (2), M m N mg Mg N N a1 a2 a1 + a2 = m 1 M 2 F m F M F 2 …(3)
Relative acceleration between M and m = ar = a1+a2. Let the block m strikes the pulley after a time t. Since, 2 r t a 2 1 m 1 M 2 F 2 t t (2m2MmM)F .
Problem 4: A piece of uniform string of mass M hangs, vertically so that its free end just touches horizontal surface of a table. If the upper end of the string is released, find at any instant, the total force on the surface just before the string falls completely.
Solution : Considering the time when y fraction of total length has fallen on the surface.
the impact force = Fimpact = v(dm/dt)
y l M dt d v = l Mv dt dy l Mv 2 ,
where y = distance fallen,
l - y v
F ' mg Fim
By putting v2 = 2gy, we obtain Finput = 2Mg (y/l)
The weight of the portion lying on the surface = m'g = (M/l)yg
Total force offered by the surface = F' = 3Mg(y/l)
Just before the falls completely, y = l F' = 3Mg.
Problem 5: A wooden box of mass 20 kg. is kept on a rough horizontal surface. The co-efficient of static friction s = 0.41 and co-efficient of kinetic friction k = 0.3 between the box and the horizontal surface. A constant force F of 70 N is applied to the box at an angle of 30o with the horizontal. Find the acceleration of the box on the surface and the frictional force on the box. Take g = 10 m/s2.
F = 70 N
M = 20 Kg
300
Solution: The force components acting on the box with respect to the chosen X-Y axes is shown in Figure. w.r.t. Y axis Mg Sin F N N=Mg-Fsin …(1) Now,fs =sN …(2) Y X f FCos f Mg N N Fsin =
]
x
70
10
x
20
[
x
41
.
12 = 0.41 x 165 = 67.65N Again F cos = 3 35 2 3 x 70 = 60.62N F cos fs N FSin f FCos MgHence decide whether the box will move or not and hence find what should be the frictional force.
Problem 6: A plank of mass M is placed on a horizontal
smooth surface. Another bar of mass m is placed on the plank. The co-efficient of static friction between the plank and the bar is µ. What max. horizontal force F on the plank will cause no slipping of the bar on the plank?
M F
m
Solution: The forces on the plank and the bar is shown in Figure.
For the bar, from F.B.D., mg N2 …(1) 1 ma f … (2) N2 N1 f m f mg N2 F Mg N1
For the plank from F.B.D. 2 Ma f F …(3) Mg N N1 2 …(4)
Again we know, maximum static frictional force on ‘m’ is 2 max N f , from (1) we have mg f N2 a2 Mg F N1 N2 f a2 i.e. fmax mg …(5)
Now to have no slipping of the bar on the plank we need .) let ( a a a1 2
from (2) and (3) we have
ma f …(6) Ma f F …(7) i.e. m f M f F i.e. m M mF f …(8)
Again for no slipping of m on M,
max f f …(9) i.e. mg m M mF . From (5) and (8) i.e. F(Mm)g …(10) i.e. Fmax (Mm)g .
Problem 7: In the above problem. if the force F would have
been applied on the bar of mass m, then what would have been the maximum force F for no slipping of m on M? The horizontal surface on which the plank is kept is smooth.
M m F
Solution: The force diagram and the F.B.D. are shown in Figure respectively. The ground is taken as a frame of ref.
With reference to F.B.D. of the bar, mg N2 …(1) 1 ma f F …(2) N2 N1 f m f mg N2 Mg F F
With reference to F.B.D. of the plank
N1 = N2 + Mg …(3)
f = Ma2 …(4)
Again we know fmax = µN2
i.e. fmax = µmg …(5) mg F N2 Mg f N1 N2 a1 a2 f
If the bar is not to slip over plank,
a1 = a2 = a (let) and ffmax …(6)
Then (2) and (4) become
F f = ma …(7)
f = Ma …(8)
Using equations (6), (7) & (8), Fmax can be found out as
Fmax= mg M m 1 .
Problem 8: Shown in the diagram is a system of two
bodies. (i) a block of mass m (ii) a disc of mass M, is held in equilibrium. If the string 3 is burnt, find the acceleration of the disc. Neglect the friction between all contacting surfaces & the mass of the
pulley P & Q. M
P
3 m Q 1 2
Solution : Referring to the free-body diagram we write the equation of motion. For the block m :
T1N = ma1 … (I)
& N – mg = 0 … (ii)
pulley P :
mpg + T2 – 2T1 = mpap Since the pulley is light
mp 0
T2 = 2T1 . . .(iii)
Now we have four equations & five unknown T1, T2, a1 a2 & N Therefore we seek another equation from the geometry of the figure the positions of the particles m, P & M are related as
x1 + 2y1 = 1 & (y2 – y1) + y2 = 2 where 1 & 2 are the lengths of the strings 1 & 2 respectively Eliminating y1 from these two equations we obtain,
x1 + 4y2 = 1 + 2 2 ; Differentiating both sides w.r.t. t, for two times we obtain 2 2 2 2 1 2
dt
y
d
4
dt
x
d
= 0 ;Differentiation of 1 and 2 are zero because 1 & 2 are constants.
a1 = 4a2 . . . (v)
(i) & (ii) yields
T1 - mg = ma1 . . . (vi)
(iii) and (iv) yields
Mg – 4T1 = a2 . . .. (vii)
Eliminating T1 from (vi) & (vii) and putting a1 = 4a2, we obtain a2 = m 16 M m 4 M g.
Problem 9: A bead of mass 'm' is fitted on to a rod and can move on it without friction. At the initial moment the bead is in the middle of the rod. The rod moves translationally in a horizontal plane with an acceleration 'a' in a direction forming an angle with the rod. Find the acceleration of the bead relative to the rod.
a
Solution : Let ar be the acceleration of the bead relative to the rod. Then ar cos is the leftward acceleration of the bead relative to the rod and ar sin is downward relative acceleration of the rod. If ay and ax be the absolute leftward horizontal and downward vertical acceleration of the bead
ar cos = ay - a
(
Relative acceleration is simply the vector difference between the absolute acceleration)or ay = ar cos + a . . . (i) and ar sin = ax 0
or ax = ar sin . . . (ii)
From FBD of the bead (projecting forces vertically and horizontally) mg N cos = m ar sin . . . (A)
and N sin = m(arcos + a) . . . (B) eliminating N between (A) and (B)
mg sin = mar + ma cos or ar = g sin a cos
Problem 10: Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a light and inextensible string and a frictionless pulley as shown in the figure. The wedge is inclined at 45o to the horizontal on both sides. The coefficient of friction between the block A and the wedge is 2/3 and that between the block B and the wedge is 1/3. If the system of A and B is released from rest, then find,
45o 45o B A 2m m
(i) the acceleration of A (ii) tension in the string
(iii) the magnitude and direction of the frictional force acting on A.
Solution : (i) In the absence of friction the block B will move down the plane and the
block A will move up the plane. Frictional force opposes this motion. F.B.D. of the blocks N1 a 2mgsin45o mgsin45o f2 f1 (B) (A) T N2 T 2mgcos45o mgcos45o a T mgsin45o f 1 = ma . . . (1) and 2 mg sin45o f 2 T = 2ma . . . (2)
Adding (1) and (2), we get mg sin45o (f
1 + f2) = 3ma
for a to be non zero mgsin45o must be greater than the maximum value of (f1 + f2) (f1 + f2)mqx = (1m1 + 22m2)gcos45o = 3 4 mgcos45o mg sin45o < (f 1 + f2)max
Hence blocks will remain stationary (ii) F.B.D. of the block B
f(2)max = mg 2 3 2 45 cos mg 2 3 1 o W| | = 2 mg sin45o = 2 mg 2 2mgsin45o f2 T N2 2mgcos45o 2 mg sin45o > f
2(max) , therefore block B has tendency to slide down the plane.
For block B to be at rest T + f2(max) = 2 mg cos45o T = 2 3 mg 4 3 2 2 2 mg
T = mg 3 2 2 (iii) mg cos45o = 2 mg
T(tension) is greater than mgcos45o.
Hence block A has tendency to move up the plane, therefore frictional force on the block A will be down the plane.
F.B.D. of A For A to be at rest mg sin45o + f = T f = T mg sin45o = 2 mg mg 3 2 2 f = 2 3 mg mgsin45 o f N T mgcos45o
8.2 Objective
Problem 1: A uniform rope is moving with a constant acceleration ‘a’ on a smooth horizontal surface. The ratio of the tension in the rope at its mid point to the applied force is,
F
(A) 1:1 (B) 1:2
(C) 0 (D) 2:1
Solution : Let total mass of the rope be M and total length L. Now let us find the tension at a point at a distance from the end where the force F (let) is applied. (L-) T T (M – m)g N1 mg N2 F a
From F.B.D. of front part shown
N2 = . g L M …(1) F T = . a L M …(2) N1 mg F =(M/L)g T a
From F.B.D. of rear part shown
N1 = L . g M M …(3) T = . a L M M …(4) (M-m)g=(M-(m/L))g N1 T a From (2) and (4), F = Ma …(5) Therefore, a = M F …(6) From (4) and (6), T = .MF L 1 M Therefore, L 1 F T …(7)
At the mid point,
2 1 L , therefore 2 1 F T
Note: 1. If it is found by mistake
T F
then it will come as T F
= 1 2 2. If the mass of the rope is not accounted then
1 1 F T Therefore (B).
Problem2: Two particles A and B, each of mass m, are inter-connected by an inextensible such that the particle B hangs below a table as shown in the figure and particle A is on a rough rotating disc at a distance r from the axis of rotation of the disc. If the angular speed of the disc is = g/r , the frictional force developed at the interface of the particle & the disc is equal to
= (g/r)1/2 r A B m m (A) mg/2 (B) < mg/2 (C) mg/2 (D) zero
Solution : The particle of mass m experiences two forces (i) tension T (ii) frictional force f. Since the particle A is rotating in a circular path of radius r, its centripetal acceleration, r 2 = m f T f T m B r A m T
Putting T = mg for equilibrium of the mass B & 2 = g/r we obtain f = mg – mr g/r = 0
Problem 3: A U shaped smooth wire has a semi-circular
bending between A and B as shown in the figure 10.3. A bead of mass ‘m’ moving with uniform speed v through the wire enters the semicircular bend at A and leaves at B. The average force exerted by the bead on the part AB of the wire is,
A B v m d (A) 0 (B) πd 2 4mv (C) πd 2 2mv (D) none of these.
Solution: Choosing the positive X-Y axis as shown in the figure, the momentum of the bead at A is pi mv. The momentum of the bead at B is pf mv.
Therefore, the magnitude of the change in momentum between A and B is
v m 2 p p p f i
i.e.p = 2mv along positive X-axis.
A B +mv = pi +y -x +x -mv = pf
The time interval taken by the bead to reach from A to B is t = v 2 / d . = v 2 d .
Fav = t p = v 2 d mv 2 = d mv 4 2 Therefore (B).
Note : 1. By mistake, if consider the change in the magnitude of the momentum will be equal to zero. Then the change in momentum he will attain wrong choice (a).
2. If someone accounts carelessly d = r instead r = d/2 then he will lead to wrong choice (c).
Problem 4 : A uniform chain is just at rest over a rough horizontal table with its (1/n) th part of length hanging vertically. The co-efficient of static friction between the chain and the table is,
(A) μ 11η (B) μ 11η x L (C) μ 1ηη (D) μ 1ηη Solution: We see that a portion of the chain is lying
on the table top. Let the mass of that portion be m1. Let the mass of the remaining (hanging) portion of the chain be m2. Since the chain is at the point of slipping, the weight of the hanging portion of the chain counterbalances the maximum static frictional force fmax between m1 and the surface.
m2 m1 f N1 m1g m2g = fmax ; m2g = N1
where N1 – m1g = 0 for the equilibrium of the portion of chain lying as on the table. m2g – m1g = 0 = 1 2 m m = ) x L ( L M x L M = 1 x/L L / x = 1 Therefore (C).
Note: If the direction of frictional force on the part of the chain lying on the table is not properly considered and length of the portions is altered contrary to the given statements, you will lead to wrong answer.
Problem 5: Two identical particles A and B, each of mass m, are interconnected by
elongation of the relative spring is x, the relative acceleration between the particles is equal to (A) F/2m (B) m kx F (C) m 2kx F (D) m kx
Solution : Equation of motion for A : T = ma a = m kx m T for B : F – T = ma a = m kx F A B F m T T m
The relative acceleration =ar = | a a | =
m kx 2
F .
Problem 6: A pendulum is hanging from the ceiling of a cage. If the cage moves up with constant acceleration a, its tension is T1 and if it moves down with same acceleration, the corresponding tension is T2. The tension in the string if the cage moves horizontally with same acceleration a is, (A) 2 2 2 T 2 1 T (B) 2 2 2 T 2 1 T (C) 2 2 2 T 2 1 T (D) 2 2 2 T 2 1 T
Solution : Referring to the free body diagrams of the bob we obtain, T1 – mg = ma T1 = m (g + a) …(1) mg T2 = ma T2 = m(g a) …(2) T1 mg a T2 mg a T mg a
When the cage moves horizontally with an acceleration a, let the tension be T. From the free body diagram, T Sin = ma
And T Cos mg = 0
(T Sin)2 + (T Cos)2 = (ma)2 + (mg)2 T2 = m2 (g2+a2) …(3) From (1) and (2) 2 2 2 2 2 1 (g a) (g a) m T m T
(
g
a
)
2
T
T
2 2 2 2 2 1
m2 …(4)Equations (3) and (4), we obtain ) T T ( 2 1 T 2 2 2 1 2 2 T T T 2 2 2 1 Therefore (C).
Problem 7: In the figure the block of mass M is at rest on the floor. The acceleration with which should a boy of mass m climb along the rope of negligible mass so as to lift the block from the floor is,
(A) = 1 g m M (B) 1 m M M a m (C) g m M (D) g m M
Solution : Equation of motion for M: Since M is stationary,
T – Mg = 0
T = Mg …(1)
Since the boy moves up with an acceleration ‘a’, T T T – mg = ma
T = m(g+a) …(2)
Equating (1) and (2), we obtain Mg = m(g+a) a = 1 g m M Mg T a That means, if a > 1 g m M
, the block M can be lifted. Therefore (B).
Problem 8 : A particle is constrained to move along a straight rod. The rod moves with a constant angular acceleration . Disregarding greater of the particle, the time after its motion begins, it stops sliding along the rod is equal to (A) (B) 1 (C) (D) Infinitesimal
Solution : If N is the normal contact force experienced by the particle, the maximum frictional force f = N. (Weight of the particle = 0 as g = 0) Since this frictional force imparts necessary centripetal force mr2 that is f = mr2 , we obtain N = mr2
N
N
By putting N = ma = m (r ) and = t, we obtain
(mr) = mr2 t2 t =
That means after
the particle will slide.
Problem 9: Two identical small masses each of mass ‘m’ are connected by a light inextensible string on a smooth horizontal floor. A constant force F is applied at the mid point of the string as shown in the figure. The acceleration of each mass towards each other is,
(A) a1 = 2 F3m (B)a1 23mF F m m (C) mF 3 2 1 a (c) (D) None of these
Solution : Let the tension in the string be T at any angular position , the acceleration of each ball along x and y axes be ‘a’ and ‘a1’ respectively. Writing the equation of motion of m, we obtain.
Fx = ma
T Cos = ma …(1)
Fy = ma1
T Sin = ma1 …(2)
At point p, as it is accelerating with an acceleration a, therefore
F – 2 T Cos = mp a where mp = mass of the string at the point P 0
F = 2 T Cos …(3) (2) (1) tan = a a1 a1 = a tan where a = m Cos T from (1). a a1 F P T a T a a1
Putting T = 2CosF from (3), we obtain. tan m 2 F a1 ,Putting = 30° m 3 2 F a1 Therefore (A).
Problem 10: A block of mass ‘m’ is kept over the smooth surface of the plank of mass M. The plank of length is kept over the smooth horizontal surface. A constant horizontal force F is applied onto the plank as shown in figure. The time after which the block falls off the plank is m M F (A) F M 2 (B) F m) (M 2 (C) F m 2 (D) none of these
Solution: F.B.D. of the block and the plank are shown in figures respectively. Since no horizontal force acts on the block it remains at rest. For vertical equilibrium of the block
For block, N1 = mg …(1)
For Plank N2 = N1+Mg …(2)
Since the horizontal force F is acting on the plank, its acceleration a is given as
F = Ma …(3) Therefore, a = F/M …(4) N1 mg Mg N2 N1 F a
The relative acceleration between the plank and the block is
ar = a-0 = F/M …(5)
The relative displacement
xr = …(6)
Therefore
= a t ,
2
1 2
r where t is the time of separation of the block from the plank
Therefore t = r a l 2 = F M 2 . Therefore (A).
Note : 1. One may write a = Force/ total mass = F/(M+m) leading to the wrong
answer (b).
