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Engineering Mechanics Test 2
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Engineering Mechanics Test 2
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  • Question 1/25
    2 / -0.33

    A car crashes against a wall. The initial velocity at collision is 15 m/sec and the velocity after collision is 2.6 m/sec in the opposite direction. The mass of the car is 1500 kg. What is the average force exerted on the automobile bumper if collision lasts for 0.15 seconds.
    Solutions

    Concept:

    Impulse (I): It is defined as the integral of force with respect to time. It is a vector quantity. Or it is also defined as a change in the linear moment (P) with respect to time.

    I = F  × dt = ΔP 

    According to Newton's second law, the Force can be defined as a moment per time.

      F=d(mv)dt

    This is known as the Impulse momentum equation.

    where, m = mass of body, v = relative velocity

    Calculation:

    Given:

    m = 1500 kg, v1 = 15 m/s, v2 = 2.6 m/s, t = 0.15 s

    Therefore, F=1500×(15(2.6))0.15=1500×17.60.15=1.76×105 N

  • Question 2/25
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    If the coefficient of kinetic friction between block B and floor is 0.2, what is acceleration of system, in m/s2, as shown in figure, is _____ (Take g = 10 m/s2 , B = 1500 N and A = 1000 N)

    Solutions

    Concept:

    FBD of Body A

     

    B = 1500 N

    A = 1000 N

    ∑Fy = ma

    1000T=100010a     ---(1)

    FBD of Body B,

    ∑Fx = ma

    TμN=150010a      ---(2)

    ∑Fy = 0

    ∴ N = W = 1500 N

    Now,
    Using equation (1) and (2)

    1000 – T = 100 a

    And T – (0.2 × 1500) = 150 a

    Adding,

    1000 – 300 = 250 a

    700250=a

    ∴ a = 2.8 m/s2

  • Question 3/25
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    Two masses are attached to the two ends of a cable which passes over two frictionless pulleys as shown in the figure. (acceleration due to gravity = g)

    The acceleration of the 100 kg mass is

    Solutions

    Explanation:

    Consider the dynamic equilibrium of 100 kg & 50 kg

    By applying de-Alembert’s principle

    T + 100 a = 100 g

    ∴ T = 100 (g - a)      ----- (1)

    T = 50g + 50 a

    T = 50 (g + a)      ----- (2)

    Equation (1) & (2)

    50(g + a) = 100 (g - a)

    g + a = 2g – 2a

    3a = g

    ∴ a = g/3

  • Question 4/25
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    For the pulley system shown in the figure below, which of the following statements are true?

    Solutions

    Concept:

    Draw the free body diagram for the pulley connected to the hinge support A

    From the horizontal equilibrium about A

    RH = P

    From the vertical equilibrium about A

    RV = P

  • Question 5/25
    2 / -0.33

    A block weighing 2500 N rests on a horizontal plane for which coefficient of friction is 0.2. This block is pulled by a force of 1000 N acting at an angle of 30° to the horizontal. Find the velocity (in m/s) of the block after it moves 30 m starting from rest.
    Solutions

    ∑Fy = 0

    W = N + P sin 30

    2500 = N + 1000 sin 30 ⇒ N = 2000 N

    F = μN = 0.2 × 2000 = 400 N

    Work done = Kinetic energy

    (Pcos30F)s=12.Wg(v2u2)(1000cos30400)×30=12.25009.81(v20)  

    v = 10.47 m/s
  • Question 6/25
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    Force in the member BD is kN (Considering Compressive as positive and tensile as negative and correct upto 1 decimal.)

    Solutions

    Since the truss in given figure has symmetric loading therefore the vertical reactions at A and F will be same and is 500 N.

    RA = RF = 500 N

    Now let us cut the truck through section X-X as shown below:

    Now taking

    ΣFv = 0

    FBD sin θ + 500 – RF = 0

    FBD sin θ + 500 – 500 = 0

    FBD = 0

    Therefore FBD = 0

  • Question 7/25
    2 / -0.33

    A bullet of mass 1 kg if fired with a velocity of u m/s from a gun of mass 10 kg. The ratio of kinetic energies of bullet and gun is
    Solutions

    Momentum conservation:

    mu = MV

    Vu=mM

    The ratio of kinetic energies of bullet and gun is:

    K.EbK.Eg=12mu212MV2=mu2MV2=mM(Mm)2=Mm=10

  • Question 8/25
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    A solid cylinder of mass m and radius r starts rolling form rest along an inclined plane. If it rolls without slipping form a vertical height h, the velocity of its centre of mass when it reaches the bottom is ________.

    Solutions

    In rolling without slipping through the distance L down the incline, the height of the rolling object changes by "h". Hence the gravitational potential energy changes by mgh.

    P.E = K.Etranslation + KErotational

    mgh=12mv2+12Iω2=12mv2+12I(vr)2

    2mgh=mv2(1+Imr2)v=2gh(1+Imr2)

    Hollow Cylinder

    I=mr2

    v=gh

    Solid Cylinder

    I=12mr2

    v=43gh

    Hollow Sphere

    I=23mr2

    v=65gh

    Solid Sphere

    I=25mr2

    v=107gh

  • Question 9/25
    2 / -0.33

    The moment of inertia of a hollow circular section whose external diameter is 8 cm and internal diameter is 6 cm, about centroidal axis is.
    Solutions

    Concept:

    Moment of inertia is the sum of the product of mass of each particle with the square of its distance from the axis of the rotation.

    M.O.I for hollow circular cross-section =π64(d04di4) 

    Where d0 = outer diameter of cross section

    di = inner diameter of cross section

    Calculation:

    d0 = 8 cm

    di = 6 cm

    I=π64(8464)

    = 137.45 cm4

    Other Important Points:

    Formula of moment of inertia for various other figures is given below.

    S.No.

    Shape of cross-section

    INA

    Ymax

    Z

    1

    Rectangle

    I=bd312

    Ymax=d2

    Z=bd26

    2

    Circular

    I=π64D4

    Ymax=d2

    Z=π32D3

    3

    Triangular

    I=Bh336

    Ymax=2H3

    Z=BH324

  • Question 10/25
    2 / -0.33

    The principle of virtual work states that, for a body to be in equilibrium, the virtual work should be:
    Solutions

    Virtual work is the work done by a real force acting through a virtual displacement or a virtual force acting through a real displacement.

    A virtual displacement is any displacement consistent with the constraints ofthe structure, i.e., that satisfy the boundary conditions at the supports.

    A virtual force is any system of forces in equilibrium.

    The principle of virtual work states, for bodies in equilibrium, for a small arbitrary displacement, the total work done by the system is zero.

    In this method, the system is displaced through a small amount about a reference point and the work done by all the forces about the reference point is summed to zero to find the unknown reactions if any.

    ∴ The principle of virtual work states that, for a body to be in equilibrium, the virtual work should be zero.
  • Question 11/25
    2 / -0.33

    A flywheel of mass 8 tonnes starts from rest and gets up a speed of 180 r.p.m. in 3 minutes. Find the torque (in Nm) exerted on it, if the radius of gyration of the flywheel is 60 cm.
    Solutions

    m = 8t = 8000 kg

    ω0 = 0 (because it starts from rest)

    ω=180r.p.m.=180×2π60=6πrad/s

    t = 3 min = 180 s

    I = Mk2 = 8000 × (0.6)2 = 2880 kg-m2

    ω = ω0 + αt

    6π = 0 + α × 180

    α=6π180=0.105rad/s2

    T = Iα = 2880 × 0.105 = 302.4 N-m
  • Question 12/25
    2 / -0.33

    A ladder 5 m long, weighing 100 N is resting against smooth wall and a rough horizontal floor whose coefficient of friction is 0.3. A person weighing 500 N is climbing along the ladder and the ladder makes an angle of 30° with floor, then which of the following statements are true?
    Solutions

    Concept:

    The ladder slips when the equilibrium is lost. Let us assume the ladder is at equilibrium till the person climbs x m from the bottom

    Calculation:

    From horizontal equilibrium

    N2 = fs    - - - (1)

    From vertical equilibrium

    N1 = 100 + 500 = 600 N

    fs = μN1 = 0.3× 600 = 180 N

    From (1),

    N2 = 180 N

    Now,

    Taking moment about point A

    (500×32×x)+(100×32×2.5)=(180×52)

    x = 0.539 m = 53.92 cm

  • Question 13/25
    2 / -0.33

    A cantilever truss is loaded as shown in figure. Find the value W (in kN), which would produce the force of magnitude 15 kN in the member AB.

    Solutions

    According to method of section:

    ME=0FAB×2=W(1.5)+3W(4.5)=15W

    FAB = 7.5 W

    7.5 W = 15 kN

    W = 2 kN

  • Question 14/25
    2 / -0.33

    Find the forces in AC and FH.

    Solutions

      

    MB=018×4+18×8=Rj×16

    Rj = 13.5 kN

    Fy=0RB=18+8Rj=22.5kN

    FBD of joint B:

    Fx=0FBD=0Fy=0FAB=22.5kN

    FBD of joint A:

    Fx=0FADcosθ+FAC=0Fy=0FADsinθ=22.5FAD(CDAD)=22.5=FAD(35)FAD=22.5×53=37.5FAD=FAD×cosθ=FAD(ACAD)FAC=37.5(45)FAC=30 kN

    MB=013.5(4)=FFH(3)FFH=18kNFy=0FFGsinθ=13.5FFG(EFFG)=FFG(35)FFG=13.5×53=22.5kNFy=0FFG=FFGcosθFFHFEG=22.5×(45)18=36kN

  • Question 15/25
    2 / -0.33

    Three marks A, B and C at a distance of 100 m each are made along a straight road. A car starting from rest and with uniform acceleration passes the mark A and takes 10 seconds to reach B and further 8 seconds to reach the mark C. What will be the velocity (in m/s) of car of A.

    Solutions

    As, s=ut+12at2

    For the motion of the car from A to B:

    100=uA(10)+12a(10)2

    100 = 10 uA + 50 a

    For the motion of the car from A to C:

    200=uA(18)+12a(18)2

    100 = 9 uA + 81 a       ----(2)

    By solving (1) & (2)

    uA = 8.61 m/s
  • Question 16/25
    2 / -0.33

    A man of mass 650 N dives vertically downwards into a swimming pool from a tower of height 25 meter. He was found to go down in water by x meter and then started rising. If the average resistance offered by the water against rising of man is 7300 N, then the value of x is
    Solutions

    Let initial velocity of man is “u” = 0

    V2 = u2 + 2gh

    V2 = 0 + 2 × 9.81 × 25

    V = 22.15 m/s

    Now at a velocity of 22.15 m/s man starts dives in to a swimming pool and goes up to x metre at which its velocity = 0

    So, V1 = 0

    u1 = 22.15 m/s

    a1 = -a (because body starts moving upward)

    v12=u12+2a1x

    0 = (22.15)2 + 2(-a) × X

    a=(22.15)22x=245.31x

    Now, the net force acting on the man must be equal to the product of mass and acceleration

    Fr – W = ma

    7300650=(6509.81)×245.31x

    x = 2.44m
  • Question 17/25
    2 / -0.33

    In an IPL match between MI and RCB, Chahal bowled a full toss delivery to Rohit, Rohit pulled it away for a maximum. The mass of the ball is 160 gms and the mass of the Rohit’s bat is 1.5 kg. The speed of ball is 90 kmph just before the hit and 144 kmph just after the hit. The bat swing speed just before hitting the ball is 54 kmph and the ball is in contact with the bat for 3 ms.

    (Assume hitting the ball with bat as a 1D collision)
    Solutions

    Concept:

    It is assumed hitting the ball with a bat as a 1D collision. During a collision, the momentum is conserved.

    Momentum lost by bat = Momentum gained by bat

    Mbat ubat - Mbat Vbat = Mball Vball - Mball uball

    Impulse on the ball will be change in momentum

    I = d(mv)

    Average force exerted by bat on the ball will be

    F = I/Δt

    Calculation:

    Given:

    Mball = 160 gms = 0.16 kg; Mbat = 1.5 kg; uball = 90 kmph = 25 m/s;

    Vball = 144 kmph = 40 m/s; ubat = 54 kmph = 15 m/s; Δt = 3 ms = 0.003 s

    Conservation of momentum:

    1.5 × 15 + 0.16 × 25 = 0.16 × 40 + 1.5 × Vbat

    ⇒ Vbat = 13.4 m/s = 48.24 kmph (Option 4)

    Impulse = change in momentum = Mball (uball – (Vball))

    ⇒ I = 0.16 (25 – (- 40)) = 10.4 kg m/s (Option 2)

    Average force exerted is given by

    Favg = I/Δt = 10.4/0.003 = 3.46 kN (Option3)

    Mistake point:

    While calculating impulse, directions of velocities are also needed to be considered because directions play a key role in kinematic analysis.

    If direction not considered, I = 0.16 × 15 = 2.4 kg.m/s

  • Question 18/25
    2 / -0.33

    If the spring has a stiffness k and an upstretched length l0. Determine the force P when the mechanism is in the position shown. Neglect the weight of the members.

    Solutions

    y = 2l cos θ , δy = -2l sinθ δθ

    x = l sin θ , δx = l cos θ δθ

    δU = 0;

    P δx + Fs δy = 0

    P (l cos θ δθ) + Fs (2l sinθ δθ) = 0

    P cosθ + 2 Fs sinθ = 0

    Fs = k(2l cosθ – l0)

    P = 2k tanθ (2l cosθ – l0)

  • Question 19/25
    2 / -0.33

    A semi-circular lamina of radius R = 75 mm is suspended from one of its corners, the angle made by its base diameter with the vertical will be________?
    Solutions

    Concept: 

    When a semi-circular lamina is suspended from one of its corners it will position itself in such a way that the vertical from the point of suspension will pass through its centroid.

    For semi-circular laminae, the centroid lies at a distance of 4R/3π from the base

    Calculation:

    tanθ=(4R3π)R

    ∴ θ = 23° 

  • Question 20/25
    2 / -0.33

    The Moment of Inertia of the shaded area for the figure shown below about x – x is _______ m4.

    Solutions

    (i) Moment of Inertia of traingular portion about base x – x is

    I1=3×(2)312=2m4

    (ii) Moment of inertia of rectangular portion about base x1 – x1 is

    I2=bd33=3×133=1m4

    (iii) Moment of Inertia of circular section about x – x

    I3=π64×(1.5)4=0.248m4

    Moment of Interia about x – x = I1 + I2 – I3

    = 2 + 1 – 0.248

    = 2.752 m4

  • Question 21/25
    2 / -0.33

    A block weighting 100 N is resting on a plane inclined with horizontal as shown in Fig. What horizontal force P is necessary to hold the body from sliding down the plane? (Coefficient of friction can be taken as 0.25)

    Solutions

    Concept:

    Equlibrium of a body is satisifed by the following conditions:

    ∑fx = 0

    ∑fy = 0

    ∑M = 0

    And for a distributed load 

    Calculation:

    Let us draw F.B.D of the block

    Now from equations of equilibrium

    ∑fy = 0

    N – P sin θ + W cos θ = 0

    N = P sin θ + W cos θ

    and ∑fx = 0

    W sin θ = P cos θ + μ⋅N

    W sin θ = P cos θ + μ (P sin θ + W cos θ)

    W sin θ - μ W cos θ = P cos θ + μ ⋅ P sin θ

    W sin θ - μ W cos θ = P (cos θ + μ sin θ)

    P=(WsinθμWcosθcosθ+μsinθ)=W(sinθμcosθ)(cosθ+μsinθ)   

    P=100(sin450.2×cos45)(cos45+0.25×sin45)

    ∴ P = 60 N

  • Question 22/25
    2 / -0.33

    A force f = (10î + 8ĵ - 5k̂) acts a point P(2, 5, 6). What will be the moment of the force about the point Q(3, 1, 4)?
    Solutions

    Solution

    Concept:

    Moment of a force about any point is given by

    m0=F×r

    where  r=QP=PQ

    Calculation:

    Given:

    F=(10i^+8j^5k^)P( 2, 5, 6), Q(3, 1, 4).

    r=QP=PQ

    r=PQ=QP    

    r=(2i^+5j^+6k^)(3i^+1j^+4k^)

    ∴ r=(1i^+4j^+2k^)

    m0=|i^j^k^1421085|

    m0=i^(2016)j^(520)+k^(840)

    m0=36i^+15j^48k^

  • Question 23/25
    2 / -0.33

    A Train of length 500 m is travelling in 30° north east direction with a uniform acceleration of 0.2 m/s2 and at t = 0, it has a speed of 54 kmph. At the same time (t = 0), a Benz car started from different location from rest and travelling in east direction with a uniform acceleration of 0.4 m/s2. At t = 25 sec, a passenger sitting in Benz observed the train on his left side.

    (Assume the angle is measured from east direction)

    Solutions

    Concept:

    If the person is sitting in train and observing train velocity, he will observe the train is at rest as the relative velocity of train w.r.t person is zero.

    VTP = VT – VP = VT – VT = 0

    If the person is outside the train and stationery, he will observe the actual speed of train.

    VTP = VT – VP = VT – 0 = VT

    If the person is sitting in another vehicle and observing the train, the observed train velocity will be different.

    VTP = VT – VP ⇒ VT = VTP + VP

    Hence, the vectors VT, VP and VTP constitute a triangle.

    Calculation:

    Given uT = 54 kmph = 15 m/s; uB = uP = 0; aT = 0.2 m/s2; aP = 0.4 m/s2;

    At t = 25 sec,

    VT = uT + aT t = 15 + 0.2 × 25 = 20 m/s;

    VP = uP + aP t = 0 + 0.4 × 25 = 10 m/s;

    Now both the velocities’ directions are different, let’s solve in each direction separately,

    We will assume, East direction as positive x – axis and North direction as positive Y – axis.

    VTX = 20 × cos 30° = 17.32 m/s;

    VTY = 20 × sin 30° = 10 m/s;

    VPX = 10 × cos 0° = 10 m/s

    VPY = 10 × sin 0° = 0 m/s;

    VTPX = VTX – VPX = 7.32 m/s;

    VTPY = VTY – VPY = 10 – 0 = 10 m/s;

    Since Both the east component and north component are positive, the direction of Relative velocity vector lies in 1st quadrant,

    ∴ Passenger observes the train moving in north east direction (Option 2)

    For angle,

    tanθ=VTPYVTPX=107.32=1.366 

    ⇒ θ = 53.8°

    As the angle is measured from East direction, so θ = 53.8° (Option 3)

    This problem can also be solved using Velocity triangle approach.

    Mistake points:

    The angle calculated (θ) is the angle with East direction (+ve x-axis)

    If the angle is measured from north direction, then the answer will be 90 - θ degrees.

  • Question 24/25
    2 / -0.33

    Two objects P and Q are travelling horizontally with velocity of 8 m/sec and 6 m/sec form left to right. They are separated by a distance of 15 m. The mass of the objects are 3 kg and 5 kg. If the coefficient of restitution is 0.7 what is the velocity (m/s) of P and Q after impact and when (seconds) and where (metres) will they impact with respect to initial positioning of Q. The corresponding answers are respectively
    Solutions

    Concept:

    The figure of two objects P and Q as per given data in question is,

    Conservation of linear momentum (P)

    If there is no external force acting on system then initial momentum of system is equal to final momentum is called conservation of liner momentum.

    mPUP+mQUQ=mPVP+mQVQ

    And 

    Coefficient of restitution (e)

    It is the ration of final relative velocity after collision to the initial relative velocity before collision.

    e=(VQVP)(UPUQ)

    Calculation:

    Given:

    UP = 8 m/s, UQ = 6 m/s, mP = 3 kg, mQ = 5 kg, e = 0.7

    mP × 8 + mQ × 6 = mVP + MQVR

    ⇒ 3 × 8 + 5 × 6 = 3 VP + 5 VR

    3 VP + 5 VR = 54 ------ (1)

    For Coefficient of restitution (e)

    0.7=VQVP86

    VQVP=1.4

    5VR + 3VP = 54 

    3VQ - 3VP = 1.4 × 3

    By adding both equations we will get, 

    8 VQ = 58.2

    VQ = 7.275 m/s

    VP = 8.875 m/s

  • Question 25/25
    2 / -0.33

    The motion of a particle moving along a curve is defined by y = x3- 3x + 150, starting with an initial velocity of V0=3i^14j^ m/s. If Vx is constant, then which of the following statements are true at x = 5 m?

    Solutions

    Concept:

    Vy=dydt

    ay=d2ydt

    Calculation:

    Given

    y = x3-3x+150

    From, V0=3i^14j^

    Vx = 3 m/s (Constant)

    Vy=dydt=3x2Vx3Vx

    Vy=3x2(3)3(3)

    Vy=9x29      (1)

    ay=18x      (2)

    Substitute x = 5 in (1) and (2)

    Vy = 216 m/s

    ay = 90 m/s2

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