Chapter 5 Laws of motion Solutions
Question - 31 : - A train runs along an unbanked circulartrack of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg.What provides the centripetal force required for this purpose – The engine orthe rails? What is the angle of banking required to prevent wearing out of therail?
Answer - 31 : -
Radius of the circular track, r =30 m
Speed of the train, v = 54km/h = 15 m/s
Mass of the train, m = 106 kg
Thecentripetal force is provided by the lateral thrust of the rail on the wheel.As per Newton’s third law of motion, the wheel exerts an equal and oppositeforce on the rail. This reaction force is responsible for the wear and rear ofthe rail
The angle of banking θ, isrelated to the radius (r) and speed (v) by the relation:
Therefore,the angle of banking is about 36.87°.
Question - 32 : - Ablock of mass 25 kg is raised by a 50 kg man in two different ways as shown inFig. 5.19. What is the action on the floor by the man in the two cases? If thefloor yields to a normal force of 700 N, which mode should the man adopt tolift the block without the floor yielding?
Answer - 32 : -
750N and 250 N in the respective cases; Method (b)
Mass of the block, m = 25kg
Mass of the man, M = 50 kg
Acceleration due to gravity, g = 10 m/s2
Force applied on the block, F =25 × 10 = 250 N
Weight of the man, W = 50 ×10 = 500 N
Case (a): When the man lifts the blockdirectly
Inthis case, the man applies a force in the upward direction. This increases hisapparent weight.
∴Action on the floor by the man = 250 + 500 =750 N
Case (b): When the man lifts the block usinga pulley
Inthis case, the man applies a force in the downward direction. This decreaseshis apparent weight.
∴Action on the floor by the man = 500 – 250 =250 N
Ifthe floor can yield to a normal force of 700 N, then the man should adopt thesecond method to easily lift the block by applying lesser force.
Question - 33 : - Amonkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximumtension of 600 N. In which of the following cases will the rope break: themonkey
(a) climbs up with an acceleration of 6 m s–2
(b) climbs down with an acceleration of 4 ms–2
(c) climbs up with a uniform speed of 5 m s–1
(d)falls down the rope nearly freely under gravity?
(Ignore the mass of the rope)
Fig.5.20
Answer - 33 : -
Case (a)
Mass of the monkey, m = 40kg
Acceleration due to gravity, g =10 m/s
Maximum tension that the rope canbear, Tmax = 600 N
Acceleration of the monkey, a =6 m/s2 upward
UsingNewton’s second law of motion, we can write the equation of motion as:
T – mg = ma
∴T = m(g +a)
=40 (10 + 6)
=640 N
Since T > Tmax,the rope will break in this case.
Case (b)
Acceleration of the monkey, a =4 m/s2 downward
UsingNewton’s second law of motion, we can write the equation of motion as:
mg – T = ma
∴T = m (g –a)
=40(10 – 4)
=240 N
Since T < Tmax,the rope will not break in this case.
Case (c)
The monkey is climbing with a uniform speedof 5 m/s. Therefore, its acceleration is zero, i.e., a = 0.
UsingNewton’s second law of motion, we can write the equation of motion as:
T – mg = ma
T – mg = 0
∴T = mg
=40 × 10
=400 N
Since T < Tmax,the rope will not break in this case.
Case (d)
When the monkey falls freely under gravity,its will acceleration become equal to the acceleration due to gravity,i.e., a = g
UsingNewton’s second law of motion, we can write the equation of motion as:
mg – T = mg
∴T = m(g – g)= 0
Since T < Tmax,the rope will not break in this case.
Question - 34 : - Two bodies A and B ofmasses 5 kg and 10 kg in contact with each other rest on a table against arigid wall (Fig. 5.21). The coefficient of friction between the bodies and thetable is 0.15. A force of 200 N is applied horizontally to A. Whatare (a) the reaction of the partition (b) the action-reaction forcesbetween A and B?What happens when the wall isremoved? Does the answer to (b) change, when the bodies are in motion? Ignorethe difference between μs and μk.
Fig.5.21
Answer - 34 : -
(a) Mass of body A, mA =5 kg
Mass of body B, mB =10 kg
Applied force, F = 200 N
Coefficient of friction, μs =0.15
Theforce of friction is given by the relation:
fs = μ (mA + mB)g
=0.15 (5 + 10) × 10
=1.5 × 15 = 22.5 N leftward
Netforce acting on the partition = 200 – 22.5 = 177.5 N rightward
Asper Newton’s third law of motion, the reaction force of the partition will bein the direction opposite to the net applied force.
Hence,the reaction of the partition will be 177.5 N, in the leftward direction.
(b) Force of friction on mass A:
fA = μmAg
=0.15 × 5 × 10 = 7.5 N leftward
Netforce exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward
Asper Newton’s third law of motion, an equal amount of reaction force will beexerted by mass B on mass A, i.e., 192.5 N acting leftward.
Whenthe wall is removed, the two bodies will move in the direction of the appliedforce.
Netforce acting on the moving system = 177.5 N
The equation of motion for the system ofacceleration a,can be written as:
Netforce causing mass A to move:
FA = mAa
=5 × 11.83 = 59.15 N
Netforce exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N
Thisforce will act in the direction of motion. As per Newton’s third law of motion,an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N,acting opposite to the direction of motion.
Question - 35 : - A block of mass 15 kg is placed on a longtrolley. The coefficient of static friction between the block and the trolleyis 0.18. The trolley accelerates from rest with 0.5 m s–2 for20 s and then moves with uniform velocity. Discuss the motion of the block asviewed by (a) a stationary observer on the ground, (b) an observer moving withthe trolley.
Answer - 35 : -
(a) Mass of the block, m = 15kg
Coefficient of static friction, μ =0.18
Acceleration of the trolley, a =0.5 m/s2
As per Newton’s second law of motion, theforce (F) on the block caused by the motion of the trolley is given bythe relation:
F = ma = 15 × 0.5 = 7.5N
Thisforce is acted in the direction of motion of the trolley.
Forceof static friction between the block and the trolley:
f = μmg
=0.18 × 15 × 10 = 27 N
Theforce of static friction between the block and the trolley is greater than theapplied external force. Hence, for an observer on the ground, the block willappear to be at rest.
Whenthe trolley moves with uniform velocity there will be no applied externalforce. Only the force of friction will act on the block in this situation.
(b) An observer, moving withthe trolley, has some acceleration. This is the case of non-inertial frame ofreference. The frictional force, acting on the trolley backward, is opposed bya pseudo force of the same magnitude. However, this force acts in the oppositedirection. Thus, the trolley will appear to be at rest for the observer movingwith the trolley.
Question - 36 : - Adisc revolves with a speed of 
rev/min, and has a radius of15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of therecord. If the co-efficient of friction between the coins and the record is0.15, which of the coins will revolve with the record?
Answer - 36 : -
Coinplaced at 4 cm from the centre
Mass of each coin = m
Radius of the disc, r = 15cm = 0.15 m
Frequencyof revolution, ν = rev/min 
Coefficient of friction, μ =0.15
Inthe given situation, the coin having a force of friction greater than or equalto the centripetal force provided by the rotation of the disc will revolve withthe disc. If this is not the case, then the coin will slip from the disc.
Coin placed at 4 cm:
Radius of revolution, r‘ = 4 cm= 0.04 m
Angular frequency, ω = 2πν
Frictional force, f = μmg= 0.15 × m × 10 = 1.5m N
Centripetalforce on the coin:
Fcent.
= 0.49m N
Since f > Fcent,the coin will revolve along with the record.
Coin placed at 14 cm:
Radius, 
= 14 cm = 0.14 m
Angular frequency, ω = 3.49s–1
Frictional force, f‘ = 1.5m N
Centripetalforce is given as:
Fcent.
= m × 0.14 × (3.49)2
= 1.7m N
Since f < Fcent.,the coin will slip from the surface of the record.
Youmay have seen in a circus a motorcyclist driving in vertical loops inside a‘death-well’ (a hollow spherical chamber with holes, so the spectators canwatch from outside). Explain clearly why the motorcyclist does not drop downwhen he is at the uppermost point, with no support from below. What is theminimum speed required at the uppermost position to perform a vertical loop ifthe radius of the chamber is 25 m?
Answer - 37 : -
Ina death-well, a motorcyclist does not fall at the top point of a vertical loopbecause both the force of normal reaction and the weight of the motorcyclistact downward and are balanced by the centripetal force. This situation is shownin the following figure.
The net force acting on the motorcyclist isthe sum of the normal force (FN) and the force due to gravity(Fg = mg).
The equation of motion for the centripetalacceleration ac, can be written as:
Fnet = mac
Normal reaction is provided by the speed ofthe motorcyclist. At the minimum speed (vmin), FN =0
Question - 38 : - A70 kg man stands in contact against the inner wall of a hollow cylindrical drumof radius 3 m rotating about its vertical axis with 200 rev/min. Thecoefficient of friction between the wall and his clothing is 0.15. What is theminimum rotational speed of the cylinder to enable the man to remain stuck tothe wall (without falling) when the floor is suddenly removed?
Answer - 38 : -
Mass of the man, m = 70 kg
Radius of the drum, r = 3 m
Coefficient of friction, μ =0.15
Frequencyof rotation, ν = 200 rev/min 
The necessary centripetal force required forthe rotation of the man is provided by the normal force (FN).
When the floor revolves, the man sticks tothe wall of the drum. Hence, the weight of the man (mg) acting downwardis balanced by the frictional force (f = μFN)acting upward.
Hence,the man will not fall until:
mg < f
mg < μFN = μmrω2
g < μrω2
Theminimum angular speed is given as:
A thin circular loop of radius R rotatesabout its vertical diameter with an angular frequency ω. Show thata small bead on the wire loop remains at its lowermost point for
.What is the angle made by theradius vector joining the centre to the bead with the vertical downwarddirection for 
?Neglect friction.
Answer - 39 : -
Let the radius vector joining the bead withthe centre make an angle θ, with the vertical downward direction.
OP = R = Radius of thecircle
N = Normal reaction
Therespective vertical and horizontal equations of forces can be written as:
mg = Ncosθ … (i)
mlω2 = Nsinθ … (ii)
InΔOPQ, we have:
l = Rsinθ … (iii)
Substituting equation (iii) inequation (ii), we get:
m(Rsinθ) ω2 = Nsinθ
mR ω2 = N …(iv)
Substituting equation (iv) inequation (i), we get:
mg = mR ω2 cosθ
… (v)
Sincecosθ ≤ 1, the bead will remain at its lowermost point for
, i.e., for 
For
or 
On equating equations (v) and (vi),we get:
