Engineering Mechanics MCQ Quiz - Objective Question with Answer for Engineering Mechanics - Download Free PDF

CONCEPT:

• Centripetal Acceleration (a): Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the centre of the circular path.
• The magnitude of centripetal acceleration,

\(a = \frac{{{v^2}}}{r}\)

Where v = velocity of the object and r = radius

• The direction of centripetal acceleration: It is always the same as that of change in velocity vector \(\left( {{\rm{\Delta }}\vec v} \right)\).
• When Δt decreases, the changes in angular displacement Δθ decreases. Due to which \(\left( {{\rm{\Delta }}\vec v} \right)\) becomes more and more perpendicular to \(\vec v\).
• When Δt → 0, \(\left( {{\rm{\Delta }}\vec v} \right)\) becomes perpendicular to the velocity vector. As the velocity vector of the particle at an instant acts along the tangent to the circular path.
• Therefore \(\left( {{\rm{\Delta }}\vec v} \right)\) and hence the centripetal acceleration vector acts along the radius of the circular path at that point and is directed towards the centre of the circular path.

EXPLANATION:

• If the body of the mass is revolving in the circular motion the magnitude of centrifugal force Fc acting on the particle is given by,

F = ma

\(F = \frac{mv^2}{r}\)​

• For the stable equilibrium, centrifugal force and centripetal force will be equal in magnitude and opposite in direction.

Explanation:

Elastic collision:

• In an elastic collision, both linear momentum and kinetic energy are conserved.
• In other words, the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision, and the total linear momentum of the system is conserved.
• For a two-body elastic collision, the conservation of linear momentum can be expressed as:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

Where, m₁ and m₂ are the masses of the two bodies, v_1i and v_2i are their initial velocities, v_1f and v_2f are their final velocities.

Inelastic collision:

• During an inelastic collision, two or more bodies collide and stick together, resulting in a single mass with a common velocity.
• The loss of kinetic energy is often associated with the internal forces and deformations that occur during the collision.
• In an inelastic collision, kinetic energy is not conserved.
• Unlike elastic collisions, where the total kinetic energy of the system remains constant, in inelastic collisions, some of the initial kinetic energy is transformed into other forms of energy, such as internal energy or deformation.
• For a two-body inelastic collision, the conservation of linear momentum is still valid, just as in elastic collisions.
• The conservation of linear momentum in a two-body inelastic collision can be expressed as:

m1v1i + m2v2i = (m1 + m2)v_f

Where, m1 and m2 are the masses of the two bodies, v1i and v2i are their initial velocities, v_f is the final velocity after collision.

• The total kinetic energy always decreases in an inelastic collision, the individual kinetic energy of one object might increase at the expense of others, due to the internal forces and deformations during the collision.
• However, even in such cases, the overall kinetic energy of the specific body will always be lower after the collision compared to its pre-collision value.

Concept:

Moment of inertia of a triangle with base a and height L

\({I_{base}} = \frac{{a{L^3}}}{{12}}\)

Parallel axis theorem

• The parallel axis theorem is used to find a moment of inertia about an axis that is at some distance from the centroidal axis and parallel to the centroidal.
• Suppose an axis is at distance d from the centroidal axis and parallel to the centroidal axis.

According to the parallel axis theorem, the moment of inertia about that axis is given by,

\({I_{about~axis~parallel~to~centroida~laxis}} = {I_{about~centroidal~axis}} + A{d^2}\)

Moment of inertia about the centroidal axis of the triangle

Applying parallel axis theorem we get

\({I_{xx}} = {I_{cc}} + A{\left( {\frac{L}{3}} \right)^2}\)

\({I_{cc}} = {I_{xx}} - A{\left( {\frac{L}{3}} \right)^2}\)

\(I_{cc}=\frac{{a{L^3}}}{{12}} - \frac{{aL}}{2} \times \frac{{{L^2}}}{9} = \frac{{a{L^3}}}{{36}}\)

\({I_{centroid}} = \frac{{a{L^3}}}{{36}}\)

Explanation:

Moment of inertia of an area or Second moment of area (M.I):

M.I of a body about any axis is defined as the summation of the second moment of all elementary areas about the axis.

I = Σ(A × d2)

Unit: m4 or mm4 or cm4

The following table shows the Second moment of inertia of different shapes

Shape	Figure	Moment of Inertia
Rectangle		\({{\rm{I}}_{{\rm{xx}}}} = \frac{{{\rm{b}}{{\rm{d}}^3}}}{{12}}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{{\rm{d}}{{\rm{b}}^3}}}{{12}}\)
Triangle		\({{\rm{I}}_{{\rm{xx}}}} = \frac{{{\rm{b}}{{\rm{h}}^3}}}{{36}}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{{\rm{h}}{{\rm{b}}^3}}}{{36}}\)
Circle		\({{\rm{I}}_{{\rm{xx}}}} = \frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}\)
Semicircle		\({{\rm{I}}_{{\rm{xx}}}} = {\rm{\;}}0.11{{\rm{R}}^4}\) \({{\rm{I}}_{{\rm{yy}}}} = \frac{{\rm{\pi }}}{8}{{\rm{R}}^4}\)
Quarter circle		\({{\rm{I}}_{{\rm{xx}}}} = 0.055{{\rm{R}}^4}\) \({{\rm{I}}_{{\rm{yy}}}} = 0.055{{\rm{R}}^4}\)

Concept -

When two infinite straight rods are perpendicular to each other, the curves having equal potential in their plane will form a rectangular hyperbola.

Explanation -

When two infinite straight rods are perpendicular to each other, the curves of equal potential in their plane form a rectangular hyperbola. This is a result of the electric potential created by the rods, and it leads to equipotential surfaces that resemble rectangular hyperbolas.

Hence Two infinite straight rods are perpendicular to each other. The curves having equal potential in their plane will be rectangular hyperbola.

Explanation:

• Momentum is conserved in all collisions.
• In elastic collision, kinetic energy is also conserved.
• In inelastic collision, kinetic energy is not conserved. In perfectly inelastic collision, objects stick together after collision.

Perfectly elastic collision:

If law of conservation of momentum and that of kinetic energy hold good during the collision.

Inelastic collision:

If law of conservation of momentum holds good during collision while that of kinetic energy is not.

Coefficient of restitution (e)

\(e = \frac{{Relative\;velocity\;after\;collision}}{{Relative\;velocity\;before\;collision}} = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\)

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For perfectly inelastic collision, e = 0

Concept:

The CG of a semicircular plate of  r radius, from its base, is

\(\bar y = {4r\over 3 \pi}\)

Calculation:

Given:

r = 33 cm

\(\bar y = {4r\over 3 \pi}={4\times 33\over3\times{22\over 7}}\)

y̅ = 14 cm

∴ the C.G. of a semicircular plate of 66 cm diameter, from its base, is 14 cm.

Additional Information

C.G. of the various plain lamina are shown below in the table. Here x̅  & y̅  represent the distance of C.G. from x and y-axis respectively.

Circle
Semicircle
Triangle
Cone
Rectangle
Quarter Circle
Solid hemisphere

Concept:

The friction force is given by:

f = μN

where μ is the coefficient of friction between the surfaces in contact, N is the normal force perpendicular to friction force.

Calculation:

Given:

μ = 0.1, m = 1 kg, F = 0.8 N

Now, we know that

From the FBD as shown below

Normal reaction, N = mg = 1 × 9.81 = 9.81 N

Limiting friction force between the block and the surface, f = μN =  0.1 × 9.81 = 0.98 N

But the applied force is 0.8 N which is less than the limiting friction force.

∴ The friction force for the given case is 0.8 N.

Concept:

If s = f(t)

Then, First derivative with respect to time represents the velocity

\(v=\frac{ds}{dt}\)

Acceleration is given by

\(a=\left( \frac{{{d}^{2}}S}{d{{t}^{2}}} \right)\)

Where s is the displacement

Calculation:

Given:

s = 2t3 – t2 - 1 and t = 1 sec.

\(\frac{ds}{dt}=6{{t}^{2}}-2t\)

\(\frac{{{d}^{2}}s}{d{{t}^{2}}}=12t-2\)

Concept:

Number of vertical forces:

∑F_y = T + R_N - W = 0 

Static friction force is given as,

F_s = μR_N 

Calculation:

Given:

W = 981 N, μ = 0.2, F_s = 100 N

Normal reaction:

\(\Rightarrow {R_N} = \frac{{100}}{{0.2}} = 500~N\)

Tension T

⇒ T = 981 - 500 = 481 N

Concept:

Equation of motion:

• The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering the force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move in a straight line.

There are three equations of motion:

v = u + at

v2 = u2 + 2as

\(s =ut + \frac{1}{2}at^2\)

where, v = final velocity, u = initial velocity, s = distance travelled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

Calculation:

Given:

Part-I:

When the ball will reach the highest point then the final velocity will be zero.

Initial velocity = u m/sec, final velocity = 0 m/sec, acceleration = -g m/sec²

applying 1st equation of motion

v = u + at

0 = u - gt₁

\(t_1=\frac{u}{g}\)

Part-II:

Initial velocity will be zero as the ball is at the highest point.

applying 1st equation of motion

v = u + at

3u = 0 + gt₂

\(t_2=\frac{3u}{g}\)

Therefore total time is:

t = t₁ + t₂

\(t=\frac{u}{g}+\frac{3u}{g}=\frac{4u}{g}\)

Concept:

The resting on between any frictional floor and a vertical wall will always satisfy all the static equilibrium condition i.e.

∑ F_x = ∑ F_y = ∑ M_{at any point} = 0

Calculation:

Given:

Length of ladder (AB) = 5 m, OB = 3 m

Let W will be the weight of the ladder, N_B and N_A will be support reaction, θ is the angle between ladder and floor and μ is the friction coefficient between ladder and floor.

Free body diagram of the ladder;

OA² = AB² - OB² , OA2 = 52 - 32 

OA2 = 16, OA = 4 m

From Δ OAB,

\(\cos θ = \frac{3}{5}\)

Now apply ∑ Fy = 0

N_B = W 
Now take moment about point A, which should be equal to zero

∑ MA = 0

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {W \times \frac{5}{2} \times \cos \theta } \right) = {N_B} \times 3\)

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {{N_B} \times \frac{5}{2} \times \frac{3}{5}} \right) = {N_B} \times 3\)

\(\left( {\mu \times 4} \right) + \left( {\frac{3}{2}} \right) =~3\)

\(\mu = \frac{3}{8}\)

Hence the value of the coefficient of friction between ladder and floor will be 3/8

Explanation:

Truss: 

• A framework composed of members joined at their ends to form a rigid structure.

Planar truss: 

• In these types of trusses, members lie in a single plane.
• The basic element of a planar truss is a triangle i.e. three members are joined at three points.
• For a planar truss, as there are three unknown support reactions (For a node or joint, there will be three unknown but dependent support reactions). so if
  1. m = 2j - 3 then system is statically determinate
  2. m < 2j - 3 then system is collapsed
  3. m > 2j - 3 then system is statically indeterminate
• where, m = number of members, j = total number of joints (including those attached to the supports)

Concept:

Work done = Area under force deflection diagram

\(Work ~done = \frac 12 \times Force\times deflection\)

Calculation:

Given:

Gradual loading

Force = 100 N, deflection = 100 mm = 0.1 m

\(Work ~done = \frac 12 \times Force\times deflection\)

\(Work ~done = \frac 12 \times 100\times 0.1=5~J\)

Explanation:

When the Force F is suddenly removed, then due to W, the rod is in rotating condition with angular acceleration \(\alpha\)

Thus the equation of motion:

\(\Sigma M=I_o\alpha;\;\;W \times \frac{L}{2} = I\alpha\)

\(As, I = \frac{mL^2}{3}= \frac{1}{3}\times\frac{W}{g}\times{L^2}\)

\(\therefore W \times \frac{L}{2} =\frac{WL^2}{3g}\times\alpha\)

\(\Rightarrow \alpha = \frac{{3g}}{{2L}}\)

\(\therefore Linear\;acceleration\;at\;centre = \alpha \times \frac{L}{2} = \frac{{3g}}{{2L}}\times \frac{L}{2}=\frac{{3g}}{4}\)

Also, the centre of the rod accelerates with linear acceleration a;

\(W-R=F\Rightarrow mg-R=ma\\R=mg-ma=mg-\frac{3}{4}mg=\frac{1}{4}mg=\frac{W}{4}\)