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

Explanation:

Collinear Forces

• Collinear forces act along the same line of action.

• Example: Forces on a rope in a tug of war, where all the forces are aligned along the rope.

Coplanar Concurrent Forces 

• These forces lie in the same plane and meet at a common point.

• Example: Forces on a rod resting against a wall, where normal reaction, weight, and friction act in one plane and meet at a point.

Additional InformationNon-Coplanar Concurrent Forces 

• These forces act in different planes but still intersect at a single point.

• Example: A tripod carrying a camera, where the three legs exert forces in different directions but all converge at the camera mount.

Non-Coplanar Parallel Forces → 4

• Forces that are parallel but not in the same plane.

• Example: The weight of benches in a classroom, where each bench’s weight acts vertically but at different locations and planes.

Explanation:

• The moment of inertia of an area represents how that area is distributed relative to a reference axis. It reflects the section's ability to resist bending — the larger the value, the stiffer the section.

• It can be calculated using integration methods by summing up the contributions of small elemental areas across the entire section. This is particularly useful for irregular or complex shapes where standard formulas are not applicable.

• The unit of moment of inertia depends on the units of both area and distance. If area is measured in square meters (m²), and distance in meters (m), then the unit of moment of inertia becomes meters to the fourth power (m⁴). Thus, units of area and distance both influence the unit of moment of inertia.

• Moment of inertia of an area (second moment of area) is different from mass moment of inertia, which concerns the distribution of mass and is used in rotational dynamics. Concepts like Routh’s Rule apply to mass moment of inertia, not to second moment of area.

 Additional Information Routh's Rule

• Routh’s Rule is a method used in dynamics to determine the mass moment of inertia of a rigid body about an arbitrary axis. It is particularly helpful when the body is unsymmetrical and does not align neatly with standard coordinate axes.

• It is based on the known mass moments of inertia about three mutually perpendicular axes that pass through a common point (usually the center of mass or centroid of the body).

• The rule allows engineers to calculate the moment of inertia about any axis that does not coincide with these principal axes, by combining the known moments of inertia and considering the orientation of the new axis.

• Routh’s Rule is used mainly in mechanical and aerospace engineering, where bodies rotate about complex axes (such as in gyroscopes, propellers, and unsymmetrical machinery), and helps in dynamic analysis and design of rotating systems.

Concept:

The block is acted upon by two horizontal forces:

F₁ = 80 N: F₂ = 70 N

Net Force: F_n = F₁ - F₂ = 80 - 70 = 10 N

The block is on a rough horizontal surface.

Friction force \(F_f=\mu N\)

where \(\mu =0.4\)

Force \(N=mg=20\times9.81=196.2N\)

So, \(F_f=0.4\times196.2=78.48N\)

Calculation:

• Compare F_net​ with friction:

  F_net​=10N
  F_f​=78.48N

• Since F_net < F_f the friction will completely oppose the motion — the block will not move.​

• Net acceleration = 0 m/sec²,  because friction prevents any movement.

Explanation:
The total time taken by a stone thrown vertically upwards to return to the ground can be calculated using the basic equations of motion. The time taken to reach the highest point is:

t = u / g
where
u = initial velocity = 98 m/sec
g = acceleration due to gravity = 9.8 m/sec²

t = 98 / 9.8 = 10 sec

Since the time taken to ascend and descend is the same, the total time to return to the ground is:
Total time = 2 × 10 sec = 20 sec

Concept: 

• The centroid is the geometric center of a plane figure or lamina. It is the point at which the entire area of the lamina can be assumed to be concentrated for analysis purposes.
• In uniform materials, the centroid corresponds to the center of mass or center of gravity, assuming uniform density and thickness.

Calculations:

Taking reference as the bottom part of the flange.\(A_1=(50\times10)= 500 mm^2\)

\(y_1=20+\frac{50}{2}=45mm\)

\(A_2=(100\times20)= 2000 mm^2\)

\(y_2=\frac{20}{2}=10mm\)

\(\bar y=\frac{A_1y_1+A_2y_2}{A_1+A_2}\)

\(\bar y=\frac{500\times45 + 2000\times45}{2000+500}=\frac{42500}{2500}=17mm\)

The section is symmetric about the vertical axis passing through the center of the web.

Therefore, the centroid lies at the centerline of the web horizontally.

\(\bar x =\frac{100}{2}=50mm\)

The centroid is (50 mm, 17 mm).

Concept:

• Average velocity = total displacement/ total time duration
• Equation of motion:
• v = u + at
• v² = u² + 2as
• s = ut + 1/2 at2

Calculation:

Given:

Time interval = 5 s & 1 s, Initial velocity u = 0, and, Acceleration a = 2 m/s²

​When an object starts from rest, then the total distance covered in time 1 sec and 5 sec is,

s = ut + 1/2 at2

Object is at rest, so, u = 0 m/s.

\(s_2 - s_1 = \frac12 a(t_2^2-t_1^2)\)

\(s_2 - s_1 = \frac12 \times 2(5^2-1^2)\)

s₂ - s₁ = 24 m

Total time taken, t = t₂ - t₁ = 5 - 1 = 4 sec

Average velocity = total displacement/ total time duration

Average velocity = 24/4 = 6 m/s

Average velocity between time 1 s and 5 s = 6 m/s

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:

Rate of change of velocity is known as acceleration. Its unit is m/s2. It is a vector quantity.

a = change in velocity/time

Equations of motion:

• v = u + at
• v2 – u2 = 2as
• \(s = ut + \frac{1}{2}a{t^2}\)

Calculation:

Given:

u = 36 km/h = 10 m/s; S = 125 m ; v = 54 km/h = 15 m/s,  t = ?

v2 – u2 = 2as

\(a = \frac{{{v^2} - {u^2}}}{{2s}} = \frac{{{{15}^2} - {{10}^2}}}{{2 \times 125}} = 0.5\;m/s^2\)

v = u + at

\(t = \frac{{v - u}}{a} = \frac{{15 - 10}}{0.5} = 10\;sec\)

CONCEPT:

Centripetal Acceleration (a_c): 

• Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the center of the circular path.
• The magnitude of centripetal acceleration,

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

Where v = velocity of the object and r = radius

Tangential acceleration (a_t):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and a_t = 3 m/s²

• Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

\(a = \sqrt {a_c^2 + a_t^2} \)

Centripetal Acceleration (a_c):

\(\therefore {a_c} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)

Hence, net acceleration 

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:

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:

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:

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

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

A fielder pulling his hand backward while catching a ball is an application of newton’s second law of motion.

Newton's Second Law of motion states that the rate of change of momentum of an object is proportional to the applied force in the direction of the force. ie., F=ma. Where F is the force applied, m is the mass of the body, and a is the acceleration produced.

Newton's Third Law of Motion states that 'To every action there is an equal and opposite reaction'.

A fielder pulls his hand backward; while catching a cricket ball coming with a great speed, to reduce the momentum of the ball with a little delay. According to Newton's Second Law of Motion; rate of change of momentum is directly proportional to the force applied in the direction.

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