Laws of Motion

  • Question 473

    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).

    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 
    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 474

    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?

    Coin placed at 4 cm from the centre
    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.
    Question 475

    A thin circular loop of radius rotates about its vertical diameter with an angular frequency ωShow that a small bead on the wire loop remains at its lowermost point for straight omega space less-than or slanted equal to space square root of straight g over straight R end rootWhat is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω square root of fraction numerator 2 straight g over denominator straight R end fraction end root? Neglect friction.

    Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

    OP = R = Radius of the circle
      N = Normal reaction
    The respective vertical and horizontal equations of forces can be written as, 
    mg = N Cosθ                         ... (i) 

    mlω2 = Sinθ                       … (ii)
    In ΔOPQ, we have
    Sin θ = l / R 
    R Sinθ                              … (iii)

    Substituting equation (iii) in equation (ii), we get
    m(R Sinθω2 = N Sinθ 

    mR ω2 = N                            ... (iv) 

    Substituting equation (iv) in equation (i), we get
    mg = mR ω2 Cosθ 

    Cosθ = g / Rω2                        ...(v) 

    Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω2 ≤ 1.
    i.e., for                      ω ≤ (g / R)1/2  
    For ω = (2g / R)1/2 
    rightwards double arrow      ω2 = 2g / R                  ...(vi) 
    On equating equations (v) and (vi), we get
      fraction numerator 2 straight g over denominator straight R end fraction = g / RCos θ  
     Cos θ = 1 / 2 
    ∴ θ = Cos-1(0.5)  =  600 , is the angle made by the radius vector joining the centre to the bead with the vertical downward direction. 
    Question 476

    Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

    • 100N

    • 80 N

    • 120 N

    • 150 N



    120 N

    In the vertical direction, weight are balanced by frictional forces.
    As the blocks are in equilibrium balance forces are in horizontal and vertical direction.
    For the system of blocks (A+B)
    F = N
    For block A, fA = 20 N and for block B.
    fB = fA +100 = 120 N

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