Laws of Motion

  • Question 473
    CBSEENPH11020084

    Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of,

    (a) the force on the 7th coin (counted from the bottom) due to all the coins on its top,

    (b) the force on the 7th coin by the eighth coin,

    (c) the reaction of the 6th coin on the 7th coin.

    Solution
    (a) Force on the seventh coin is exerted by the weight of the three coins on its top.
    Weight of one coin = mg
    Weight of three coins = 3mg
    Hence, the force exerted on the 7th coin by the three coins on its top is 3mg.
    This force acts vertically downward.
    (b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.
    Weight of the eighth coin = mg
    Weight of the ninth coin = mg
    Weight of the tenth coin = mg
    Total weight of these three coins = 3m
    Hence, the force exerted on the 7th coin by the eighth coin is 3mg.
    This force acts vertically downward.

    (c) The 6th coin experiences a downward force because of the weight of the four coins (7th, 8th, 9th, and 10th) on its top. 
    Therefore, the total downward force experienced by the 6thcoin is 4mg. 
    As per Newton’s third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction.
    Hence, the reaction force of the 6th coin on the 7thcoin is of magnitude 4mg.
    This force acts in the upward direction. 
    Question 474
    CBSEENPH11020085

    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?

    Solution
    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°.
    Question 475
    CBSEENPH11020086

    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?


    Solution

    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. 

    Question 476
    CBSEENPH11020087

    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).


    Solution
    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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