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A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of 6 m s–2

(b) climbs down with an acceleration of 4 m s–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).


Case (a):

Mass of the monkey, m = 40 kg

Acceleration due to gravity, g = 10 m/s 

Maximum tension that the rope can bear, Tmax = 600 N 

Acceleration of the monkey, a = 6 m/s2 , upward

Using Newton’s second law of motion, equation of motion is

                   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 

Using Newton’s second law of motion, the equation of motion is, 

                                  mg – ma 

∴                                        T = (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 speed of 5 m/s.

Therefore, its acceleration is zero, i.e., 
a = 0. 

Using Newton’s second law of motion, equation of motion is, 

                     T – m= ma 

                     T – mg = 0
 

∴                           m

                               = 40 × 10  

                               = 400 N 

Since T < Tmax, the rope will not break in this case.

Case (d) 

When the monkey falls freely under gravity, acceleration will become equal to the acceleration due to gravity, i.e., a = 

Using Newton’s second law of motion, we can write the equation of motion as:

                          mg – T = m

∴                                 T = m(g – g) = 0 

Since T < Tmax, the rope will not break in this case. 
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A train runs along an unbanked circular track 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 or the rails? What is the angle of banking required to prevent wearing out of the rail?

Given, 

Radius of the circular track, r = 30 m

Speed of the train, v = 54 km/h = 15 m/s

Mass of the train, m = 106 kg 

The centripetal 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 opposite force on the rail.

This reaction force is responsible for the wear and rear of the rail. 

The angle of banking θ, is related to the radius (r) and speed (v) by the relation:

             tan θ = v2 / rg 

                      = 152 / (30 × 10) 

                      = 225 / 300 

                   θ = tan-1 (0.75) = 36.87

Therefore, the angle of banking is about 36.87°.
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A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. 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?


Given,

Mass of the block, m = 25 kg

Mass of the man, M  = 50 kg 

Acceleration due to gravity, g = 10 m/s

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 block directly

In this case, the man applies a force in the upward direction.

This increases his apparent weight. 

Therefore,

Action on the floor by the man = 250 + 500 = 750 N


Case (b): When the man lifts the block using a pulley 

In this case, the man applies a force in the downward direction. This decreases his apparent weight. 

Therefore,

Action on the floor by the man = 500 – 250

                                               = 250 N


If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force. 

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A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

The centrifugal force experienced by the man, when the drum rotates =  mrω2 , in outward direction which is normal to the wall of cylinder. 

When the floor is removed, due to the weight of the man he has tendency to fall from the drum.

But, the man will not fall down as, 
                 μN = μmrω2, opposes it.

The man remains stuck to the wall without falling when the floor is suddenly removed if the force of friction is greater or equal to the weight of the man.

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i.e.,                 space space straight omega equals 4.67 space rad divided by straight s, is the minimum rotational speed. 
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You may have seen in a circus a motorcyclist driving in vertical loops inside a death well (a hollow spherical chamber with holes, so that the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?

The motorcyclist does not drop down when he is at the uppermost point if centrifugal force on motorcyclist is greater or equal to the weight of motorcyclist.

i.e.

              fraction numerator m v squared over denominator R end fraction greater or equal than m g 

rightwards double arrow          v greater or equal than square root of r g end root 

The minimum speed of motorcyclist not to fall from uppermost point is given by 

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Here, given

              r = 25m,            g equals 9.8 space m divided by s squared 

∴           v subscript m i n end subscript equals square root of 25 cross times 9.8 end root space equals space 15.7 space straight m divided by straight s
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