Laws of Motion

  • Question 477
    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 478
    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.
    Question 479
    CBSEENPH11020090

    The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s–2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).


    Solution
    We have, 
    Mass of the box, m = 40 kg
    Coefficient of friction, μ = 0.15
    Initial velocity, u = 0
    Acceleration, a = 2 m/s2

    Distance of the box from the end of the truck, s' = 5 m
    According to Newton's second law of motion,
    Force on the box caused by the accelerated motion, F = ma 
                   = 40 × 2
                    = 80 N 
    According to Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction.
    The backward motion of the box is opposed by the force of friction f, acting between the box and the floor of the truck.
    This force is given by, 
                     f = μmg  
                       = 0.15 × 40 × 10
                       = 60 N 
    Therefore,
    Net force acting on the block, 
                  Fnet = 80 – 60 = 20 N, acting backward 
    The backward acceleration produced in the box is, 
    Acceleration, aback = straight F subscript net over straight m space equals space 20 over 40 space equals space 0.5 space straight m divided by straight s squared space
    Using the second equation of motion,
    Time t is, 
                       s' = ut + (1/2) abackt

                       5 = 0 + (1/2) × 0.5 × t

    ∴                  t = √20 s 
    Hence, the box will fall from the truck after square root of 20 s from the start. 
    The distance s, travelled bytruck in √20 s is given by, 
                       s = ut + (1/2)at

                         = 0 + (1/2) × 2 × (square root of 20)

                         = 20 m 
    Therefore, at a distance of 20 m, the box will fall off the truck. 
    Question 480
    CBSEENPH11020091

    A disc revolves with a speed of 33 space begin inline style 1 third end style rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

    Solution
    Coin placed at 4 cm from the centre
    Given, 
    Mass of each coin = m

    Radius of the disc, r = 15 cm = 0.15 m
    Frequency of revolution, ν = 100 over 3 rev/min
                                             = fraction numerator 100 over denominator left parenthesis 3 space straight x space 60 right parenthesis end fraction
                                             = open parentheses 5 over 9 close parentheses rev/s
    Coefficient of friction, μ = 0.15
    In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.
    Case 1: Coin placed at 4 cm
    Radius of revolution, r' = 4 cm = 0.04 m
    Angular frequency, ω = 2πν
                                     = 2 x open parentheses 22 over 7 close parentheses space x space open parentheses 5 over 9 close parentheses 
                                      = 3.49 s-1

    Frictional force, f = μmg
                              = 0.15 × m × 10
                              = 1.5m N 
    Centripetal force on the coin is, 
                  Fcent. = mr'ω

                            = m × 0.04 × (3.49)

                            = 0.49 m N 
    Since f > Fcent, the coin will revolve along with the record.
    Case 2: Coin placed at 14 cm 
    Radius, r" = 14 cm = 0.14 m 
    Angular frequency, ω = 3.49 s–1 

    Frictional force, f' = 1.5 m N 
    Centripetal force is given as, 
                    Fcent. = mr"ω

                             = m × 0.14 × (3.49)2 
                             = 1.7m N 
    Since f < Fcent., the coin will slip from the surface of the record.

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