Laws of Motion

  • Question 469
    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 470
    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. 
    Question 471
    CBSEENPH11020088

    Two bodies and of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally toA. What are

    (a) the reaction of the partition

    (b) the action-reaction forces between 
    and B?

    What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ
    s and μk.


    Solution
    (a)
    Mass of body A, mA = 5 kg
    Mass of body B, mB = 10 kg
    Applied force, = 200 N
    Coefficient of friction, μs = 0.15
    The force of friction is given by the relation, 
                  fs = μ (mA + mB)g 
                     = 0.15 (5 + 10) × 10 
                     = 1.5 × 15
                     = 22.5 N, leftward 
    Net force acting on the partition = 200 – 22.5
                                                     = 177.5 N, rightward 
    According to Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.
    Hence, the reaction of the partition will be 177.5 N,directed leftwards.
    (b) 
    Force of friction on mass A, fA = μmA
                                            = 0.15 × 5 × 10
                                            = 7.5 N leftward 
    Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N, rightward 
    Now, according to Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A.
    i.e., Net force acting on mass A by mass B= 192.5 N acting leftward.
    When the wall is removed, the two bodies will move in the direction of the applied force.
    Net force acting on the moving system = 177.5 N
    The equation of motion for the system of acceleration a, can be written as,
    Net force = (mA + mB) a
    ∴ Acceleration, a = fraction numerator Net space Force over denominator straight m subscript straight A space plus space straight m subscript straight B end fraction 
                           = fraction numerator 177.5 over denominator left parenthesis 5 plus 10 right parenthesis end fraction space equals space fraction numerator 177.5 over denominator 15 end fraction space equals space 11.83 space straight m divided by straight s squared
    Net force causing mass A to move,
                     FA = mA

                         = 5 × 11.83
                         = 59.15 N 
    Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N 
    This force will act in the direction of motion.
    According to 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 472
    CBSEENPH11020089

    A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s–2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by

    (a) a stationary observer on the ground,

    (b) an observer moving with the trolley.


    Solution
    Given,
    Mass of the block, m = 15 kg
    Coefficient of static friction, μ = 0.18 
    Acceleration of the trolley, a = 0.5 m/s

    According to Newton’s second law of motion,
    Force (F) on the block caused by the motion of the trolley is given by the relation, 
                       F = ma 
                         = 15 × 0.5
                         = 7.5 N
    This force is acted in the direction of motion of the trolley. 
    Force of static friction between the block and the trolley, 
                                 f = μm
                                   = 0.18 × 15 × 10
                                   = 27 N 
    The applied external force is less than the force of static friction between the block and the trolley. Therefore, the block will appear to be at rest, for an observer on the ground. 
    When the trolley moves with uniform velocity, only the force of friction will act in this situation and there will be no applied external force. 
    (b) The person who is moving with the trolley has some acceleration. 
    The frictional force is acting backwards on the trolley and is opposed by a pseudo force of the same magnitude. But, this force is acting in the backward direction. Therefore, for an observer moving with the trolley, the trolley will appear to be at rest.

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