Engineering Mechanics Miscellaneous

Engineering Mechanics Miscellaneous

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Engineering Mechanics

Engineering Mechanics Miscellaneous
  1. A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure. Block R is tied to the wall by a mass less and in extensible string PQ. If the coefficient of static friction for all surfaces is 0.4, the minimum force F(in kN) needed to move the block S is








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    F = μ (Ws + WR) + μ WR
    F = μ (Ws + 2WR)
    = 0.4 (150 + 2 × 1000)9.81 = 1.37 kN

    Correct Option: D


    F = μ (Ws + WR) + μ WR
    F = μ (Ws + 2WR)
    = 0.4 (150 + 2 × 1000)9.81 = 1.37 kN

  1. A1 kg block is resting on a surface with coefficient of friction μ = 0.1. A force of 0.8 N is applied to the block as shown in the figure. The friction force is








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    Limiting friction force between block and the surface is 0.98 N. But applied force is 0.8 N which is less than the limiting friction force. Therefore friction force for the given case is 0.8 N.

    Correct Option: B

    Limiting friction force between block and the surface is 0.98 N. But applied force is 0.8 N which is less than the limiting friction force. Therefore friction force for the given case is 0.8 N.

  1. A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is μ = 0.2. A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right?








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    Net normal force on block
    N = W – T
    = 981 – T
    Frictional force,f = μN
    Under equilibrium, i.e. when man is just able to move the block
    μN = 100
    ⇒ μ(981–T) = 100
    ⇒ 0.2 (981 – T) = 100

    T = -
    100
    		+ 981
    0.2

    = 981 – 500 = 481N

    Correct Option: C

    Net normal force on block
    N = W – T
    = 981 – T
    Frictional force,f = μN
    Under equilibrium, i.e. when man is just able to move the block
    μN = 100
    ⇒ μ(981–T) = 100
    ⇒ 0.2 (981 – T) = 100

    T = -
    100
    		+ 981
    0.2

    = 981 – 500 = 481N

  1. A block of mass M is released from point P on a rough inclined plane with inclination angle θ, shown in the figure below. The coefficient of friction is μ. If μ < tan θ, then the time taken by the block to reach another point Q on the inclined plane, where PQ = s, is








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    Mg sin θ – μ Mg cos θ= Ma
    ⇒ a = g sin θ – μ g cos θ

    Now s = ut +
    1
    		at2
    2

    But u = 0

    Correct Option: A


    Mg sin θ – μ Mg cos θ= Ma
    ⇒ a = g sin θ – μ g cos θ

    Now s = ut +
    1
    		at2
    2

    But u = 0

  1. Two books of mass 1 kg each are kept on a table, one over the other. The coefficient of friction on every pair of contacting surfaces is 0.3. The lower book is pulled with a horizontal force F. The minimum value of F for which slip occurs between the two books is








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    [amax]A = ?
    [amax]A → [(fs)max]AB
    A → 3N
    [(fx)max]AB = μsNAB
    = 0.3 × 9.81 = 2.943 N

    (aA)max =
    [(fs)max]AB
    		= 2.943 m/s2
    mA

    At that condition,

    Fmin – 8.829 = mB aB
    Fmin = 8.829 + 1 × 2.943
    Fmin = 11.772 N

    Correct Option: B

    [amax]A = ?
    [amax]A → [(fs)max]AB
    A → 3N
    [(fx)max]AB = μsNAB
    = 0.3 × 9.81 = 2.943 N

    (aA)max =
    [(fs)max]AB
    		= 2.943 m/s2
    mA

    At that condition,

    Fmin – 8.829 = mB aB
    Fmin = 8.829 + 1 × 2.943
    Fmin = 11.772 N