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

Explanation:

For member GH,

As point G can move in a slot, hence force provided by member BE will be perpendicular to the member.

Referring to FBD,

ΣM_H = 0 

⇒ R_G = 0

⇒ Force in member GH is zero.

As member AB, CD, and EF are link members, hence they can carry only axial forces.

Referring to FBD,

​

ΣF_x = 0

⇒ F_AB = Q

ΣM_E = 0

⇒ - Q × l/2 + F_CD × l - Q × l/2 + F_AB × l = 0

Putting the value of F_AB = 0.

F_CD = 0

ΣF_y = 0

⇒ F_EF = Q

Hence member CD and GH carry zero force.

Explanation:

When two or more forces act on a body, they are called to form a system of forces.

Coplanar forces: 

• The forces, whose lines of action lie on the same plane, are known as coplanar forces.

Colinear forces: 

• The forces, whose lines of action lie on the same line, are known as colinear forces.

Concurrent forces: 

• The forces, which meet at one point, are known as concurrent forces. The concurrent forces may or may not be colinear.

Coplanar concurrent forces:

• The forces, which meet at one point and their lines of action also lie on the same plane, are known as coplanar concurrent forces.

Coplanar non-concurrent forces:

• The forces, which do not meet at one point, but their lines of action lie on the same plane, are known as coplanar non-concurrent forces.

Non-coplanar concurrent forces: 

• The forces, which meet at one point, but their lines of action do not lie on the same plane, are known as non-coplanar concurrent forces.
• A rigid body is subjected to a non-coplanar concurrent force system, \(\rm \sum F_x=\sum F_y=\sum F_z=0\)

Non-coplanar non-concurrent forces: 

• The forces, which do not meet at one point and their lines of action do not lie on the same plane, are called non-coplanar non-concurrent forces.

Explanation:

Lami's theorem is an equation relating the magnitudes of three coplanar, concurrent and non-collinear forces, which keeps an object in static equilibrium, with the angles directly opposite to the corresponding forces.

According to the theorem:

\(\frac{A}{{\sin \alpha }} = \frac{B}{{\sin\beta }} = \frac{C}{{\sin\gamma }}\)

Concept:

The moment carrying capacity (M) of any beam is given by:

M = σZ

where σ = bending stress and Z = section modulus.

Section Modulus is given by:

\(Z=\frac{I}{y_{max}}\)

where I = MOI of cross-section about NA and y = distance of extreme fiber from NA.

MOI for a solid circular is given by:

\(I=\frac{\pi{D^4}}{64}\)

MOI for a hollow circular is given by:

\(I_{hc}=\frac{\pi{(D^4\;-\;{d^4})}}{64}\)

Calculation:

Given:

ymax = D/2 for both solid and hollow shaft.

\(Z_{solid}=\frac{I}{y_{max}}\Rightarrow\frac{\frac{\pi{D^4}}{64}}{\frac{D}{2}}=\frac{\pi{D^3}}{32}\)

∴ \(Z_{hollow}=\frac{I_{hc}}{y_{max}}\Rightarrow\frac{\frac{\pi({D^4\;-\;{d^4)}}}{64}}{\frac{D}{2}}=\frac{\pi}{32}\times\frac{(D^4\;-\;{d^4})}{D}\)

Concept:

Let the length of the rod be 'L' and 'x' be the distance between the resultant force and smaller force, therefore the distance between the larger force and resultant will be (L - x)

According to the principle of moments:

"The algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”

Calculation:

Given:

F1 = 40 N, F2 = 80 N, L = 240 mm.

Since the two like parallel forces are acting on the road, so the resultant force will be the algebraic sum of the magnitude of the two forces.

Hence, the resultant force F_R is given by

F_R = F₁ + F₂ = 40 N + 80 N = 120 N

When we calculate the moment about the point of the resultant force, then the moment produced by two forces will be equal and opposite.

F₁× x = F₂ × (L - x)

40 × x = 80 × (240 - x)

40x = 19200 - 80x

120x = 9600

x = 160 mm.

∴ the distance between the smaller force and the resultant is 160 mm.

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:

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:

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:

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

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;

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

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:

• Parallel axis theorem: Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of the body about an axis parallel to given axis and passing through centre of mass of the body Io and Ma2, where ‘M’ is the mass of the body and ‘a’ is the perpendicular distance between the two axes.

⇒ I = I_o + Ma²

EXPLANATION:

• For a uniform rod with negligible thickness, the moment of inertia about its centre of mass is:

\({I_{cm}} = \frac{1}{{12}}M{L^2}\)

Where M = mass of the rod and L = length of the rod

∴ The moment of inertia about the end of the rod is

\(\Rightarrow {I_{end}} = {I_{cm}} + M{d^2} \)

\(\Rightarrow I_{end}= \frac{1}{{12}}M{L^2} + M{\left( {\frac{L}{2}} \right)^2} = \frac{1}{3}M{L^2}\)

CONCEPT:

Principle of homogeneity of dimensions:

• According to this principle, a physical equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equation are the same.
• This principle is based on the fact that only the physical quantities of the same kind can be added, subtracted, or compared.
• Thus, velocity can be added to velocity but not to force.

EXPLANATION

Given - F = at + bt²

From the principle of dimensional homogeneity, the left-hand side of the equation dimensionally equal to the right-hand side of the equation.

The dimension formula of force (F) = [MLT^-2]

∴ [MLT^-2] = [a] [T]

\(⇒ \left[ a \right] = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ T \right]}} = \left[ {ML{T^{ - 3}}} \right]\)

For second term,

⇒ [MLT^-2] = [b] [T²]

\(\Rightarrow \left[ b \right] = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{T^2}} \right]}} = \left[ {ML{T^{ - 4}}} \right]\)

Concept

The equation of motions are 

v = u + at 

v² = u² + 2as                

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

Calculation:

Given:

Mass of the block, m = 5 kg, Inclination angle of the plane, θ = 30°, Initial velocity of the block, u = 0 m/s, Distance travelled by the block, s = 3.6 m

Force acting on the block along with the inclination of plane =  \(mg\sin 30^\circ = 5 \times 10 \times \frac{1}{2}\Rightarrow 25\;N\)

Acceleration of the block along the inclined plane, a = \(g\sin 30^\circ = 5~m/{s^2}\) 

Applying equation of motion along the inclined plane.

v² = u² + 2as

v² = 0 + 2 × 5 × 3.6

∴ v = 6 m/s.

CONCEPT:

Moment of inertia:

• The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
• The moment of inertia of a particle  is

⇒ I = mr2

Where r = the perpendicular distance of the particle from the rotational axis.

• Moment of inertia of a body made up of a number of particles (discrete distribution)

⇒ I = m1r12 + m2r22 + m3r32 + m4r42 + -------

Rotational kinetic energy: 

• The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
• A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
• Mathematically rotational kinetic energy can be written as -

⇒ KE \( = \frac{1}{2}I{\omega ^2}\)

Where I = moment of inertia and ω = angular velocity.

EXPLANATION:

• The moment of inertia of the ring about an axis passing through the center and perpendicular to its plane is given by

⇒ I_ring = MR²

• Moment of inertia of the disc about an axis passing through center and perpendicular to its plane is given by -

\(⇒ {I_{disc}} = \frac{1}{2}M{R^2}\)

• As we know that mathematically rotational kinetic energy can be written as

\(⇒ KE = \frac{1}{2}I{\omega ^2}\)

• According to the question, the angular velocity of a thin disc and a thin ring is the same. Therefore, the kinetic energy depends on the moment of inertia.
• Therefore, a body having more moments of inertia will have more kinetic energy and vice - versa.
• So, from the equation, it is clear that,

⇒ Iring > Idisc

∴ Kring > Kdisc

• The ring has higher kinetic energy.