Engineering Mechanics Test 2

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.

∴  \({\bf{F}} = \frac{{{\bf{d}}\left( {{\bf{mv}}} \right)}}{{{\bf{dt}}}}\)

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, \(F=\frac{1500\times (15-(-2.6))}{0.15}=\frac{1500\times17.6}{0.15}=1.76\times10^5~N\)

Concept:

FBD of Body A

B = 1500 N

A = 1000 N

∑F_y = ma

\(1000 - T = \frac{{1000}}{{10}}a\)     ---(1)

FBD of Body B,

∑F_x = ma

\(T - \mu N = \frac{{1500}}{{10}}a\)      ---(2)

∑F_y = 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

\(\therefore \frac{{700}}{{250}} = a\)

∴ a = 2.8 m/s²

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,

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.

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)

• 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

Momentum conservation:

mu = MV

\(\frac{V}{u}=\frac{m}{M}\)

The ratio of kinetic energies of bullet and gun is:

\(\frac{K.E_b}{K.E_g}=\frac{\frac{1}{2}mu^2}{\frac{1}{2}MV^2}=\frac{mu^2}{MV^2}=\frac{m}{M}(\frac{M}{m})^2=\frac{M}{m}=10\)

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 \( = \frac{\pi }{{64}}\left( {d_0^4 - d_i^4} \right)\) 

Where d₀ = outer diameter of cross section

d_i = inner diameter of cross section

Calculation:

d₀ = 8 cm

d_i = 6 cm

\(I = \frac{\pi }{{64}}\left( {{8^4} - {6^4}} \right)\)

= 137.45 cm⁴

Other Important Points:

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

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.

m = 8t = 8000 kg

ω₀ = 0 (because it starts from rest)

\(\omega = 180\;r.p.m. = \frac{{180 \times 2\pi }}{{60}} = 6\pi \;rad/s\)

t = 3 min = 180 s

I = Mk² = 8000 × (0.6)² = 2880 kg-m²

ω = ω₀ + αt

6π = 0 + α × 180

\(\alpha = \frac{{6\pi }}{{180}} = 0.105\;rad/{s^2}\)

Concept:

Moment of inertia of circular plate,

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:

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

To find the resultant of two equal forces P making an angle θ, we use the formula for the resultant of two forces:

Here’s the derivation:
- The two forces P are equal in magnitude and form an angle θ with each other.
- The magnitude of the resultant force is given by  is the angle between one of the forces and the resultant.

The other options involve sine, tangent, or cotangent functions, which do not correctly represent the magnitude of the resultant in this context.

Moment of inertia for a circular section about an axis perpendicular to the section is given by , where d is the diameter of the circle. Therefore, the correct answer is Option C.

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)

\(v_1^2 = u_1^2 + 2{{\rm{a}}_1}{\rm{x}}\)

0 = (22.15)² + 2(-a) × X

\({\rm{a}} = \frac{{ - {{\left( {22.15} \right)}^2}}}{{ - 2{\rm{x}}}} = \frac{{245.31}}{{\rm{x}}}\)

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

F_r – W = ma

\(7300-650 = \left( {\frac{{650}}{{9.81}}} \right) \times \frac{{245.31}}{{\rm{x}}}\)

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

M_bat u_bat - M_bat V_bat = M_ball V_ball - M_ball u_ball

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:

M_ball = 160 gms = 0.16 kg; M_bat = 1.5 kg; u_ball = 90 kmph = 25 m/s;

V_ball = 144 kmph = 40 m/s; u_bat = 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 × V_bat

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

Impulse = change in momentum = M_ball (u_ball – (V_ball))

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

Average force exerted is given by

F_avg = 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

The correct answer is: B. Greater than

Explanation: For a vehicle to stay on a level circular path without skidding, the force of friction between the wheels and the ground must be greater than or equal to the centrifugal force. The frictional force provides the necessary centripetal force that keeps the vehicle in circular motion. If the frictional force is less than the centrifugal force, the vehicle will skid out of the circular path.

The problem asks for the angle between two forces when the resultant is maximum and minimum.

Solution: When two forces are applied at an angle:

• The resultant force is maximum when the angle between them is 0°. This happens because the forces add up completely in the same direction.

• The resultant force is minimum when the angle between the two forces is 180°, as the forces are in opposite directions and subtract from each other.

Therefore, the correct answer is: A. 0° and 180°.

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

∑f_y = 0

N – P sin θ + W cos θ = 0

N = P sin θ + W cos θ

and ∑f_x = 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)

\(\Rightarrow P=\left( \frac{W\sin \theta -\mu W\cos \theta }{\cos \theta +\mu \sin \theta } \right)=W\frac{(\sin \theta -\mu \cos \theta )}{(\cos \theta +\mu \sin \theta )}\)   

\(\Rightarrow P=100\frac{(\sin 45-0.2\times \cos 45)}{(\cos 45+0.25\times \sin 45)}\)

∴ P = 60 N

The correct answer is: D. All of the above

Efficiency of a lifting machine can be expressed in different ways, all of which are related:

• Output to the input: This refers to the ratio of useful work (output) to the total work (input) done on the machine.

• Work done by the machine to the work done on the machine: This is essentially the same as the above, representing the ratio of the machine's work output to the input work.

• Mechanical advantage to the velocity ratio: Efficiency can also be defined as the ratio of mechanical advantage (MA) to velocity ratio (VR), which is a measure of how effectively the machine converts input energy to output work.

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.

V_TP = V_T – V_P = V_T – V_T = 0

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

V_TP = V_T – V_P = V_T – 0 = V_T

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

V_TP = V_T – V_P ⇒ V_T = V_TP + V_P

Hence, the vectors V_T, V_P and V_TP constitute a triangle.

Calculation:

Given u_T = 54 kmph = 15 m/s; u_B = u_P = 0; a_T = 0.2 m/s²; a_P = 0.4 m/s²;

At t = 25 sec,

V_T = u_T + a_T t = 15 + 0.2 × 25 = 20 m/s;

V_P = u_P + a_P 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.

V_TX = 20 × cos 30° = 17.32 m/s;

V_TY = 20 × sin 30° = 10 m/s;

V_PX = 10 × cos 0° = 10 m/s

V_PY = 10 × sin 0° = 0 m/s;

V_TPX = V_TX – V_PX = 7.32 m/s;

V_TPY = V_TY – V_PY = 10 – 0 = 10 m/s;

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

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

For angle,

\(\tan \theta = \frac{{{V_{TPY}}}}{{{V_{TPX}}}} = \frac{{10}}{{7.32}} = 1.366\) 

⇒ 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.

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.

\({{\rm{m}}_P}{\rm{\;}}{{\rm{U}}_P} + {\rm{\;}}{{\rm{m}}_Q}{\rm{\;}}{{\rm{U}}_Q} = {{\rm{m}}_P}{\rm{\;}}{{\rm{V}}_P} + {{\rm{m}}_Q}{\rm{\;}}{{\rm{V}}_Q}\)

And 

Coefficient of restitution (e)

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

\({\rm{e}} = \frac{{\left( {{{\rm{V}}_Q} - {{\rm{V}}_P}} \right)}}{{\left( {{{\rm{U}}_P} - {{\rm{U}}_Q}} \right)}}\)

Calculation:

Given:

U_P = 8 m/s, U_Q = 6 m/s, m_P = 3 kg, m_Q = 5 kg, e = 0.7

m_P × 8 + m_Q × 6 = m_P­V_P + M_QV_R

⇒ 3 × 8 + 5 × 6 = 3 V_P + 5 V_R

3 V_P + 5 V_R = 54 ------ (1)

For Coefficient of restitution (e)

\(0.7 = \frac{{{V_Q} - {V_P}}}{{8 - 6}}\)

\({V_Q} - {V_P} = 1.4\)

5V_R + 3V_P = 54 

3V_Q - 3V_P = 1.4 × 3

By adding both equations we will get, 

8 V_Q = 58.2

∴ V_Q = 7.275 m/s

V_P = 8.875 m/s

Concept:

\({V_y} = \frac{{dy}}{{dt}}\)

\({a_y} = \frac{{{d^2}y}}{{dt}}\)

Calculation:

Given

y = x³-3x+150

From, \(\overrightarrow {{V_0}} = 3\hat i - 14\hat j\)

V_x = 3 m/s (Constant)

\({V_y} = \frac{{dy}}{{dt}} = 3{x^2}{V_x} - 3{V_x}\)

\({V_y} = 3{x^2}\left( 3 \right) - 3\left( 3 \right)\)

\({V_y} = 9{x^2} - 9\)      (1)

\({a_y} = 18x\)      (2)

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

V_y = 216 m/s

a_y = 90 m/s²