Engineering Mechanics - 1 - Mechanical Engineering MCQ

Two parallel forces equal in magnitude and opposite in direction and separated by a definite distance are said to form a couple.

The translatory effect of a couple on the body is zero.

The only effect of a couple is a moment and this moment is same about any point, the effect of a couple is unchanged if:

(i) The couple is rotated through any plane

(ii) The couple is shifted to any other position

(iii) The couple is replaced by another pair of forces whose rotational effect is same

u_A = u_B = 54 Km/hr = 15 m/s; a_A = 6 m/s², a_B = 0

Car B is 300 m ahead of car A.

Let both move in time t sec.

Distance travelled by A in time t = 300 + Distance travelled by B in time t

A couple consists of two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. It does not produce any translation, only rotation. The resultant force of a couple is zero. But, the resultant of a couple is not zero; it is a pure moment.

Conservation of linear  momentum:

Collinear Forces: Line of action of all the forces act along the same line.

Coplanar parallel forces: All forces are parallel to each other and lie in a single plane

Coplanar concurrent forces: Line of action of all forces pass through a single point and forces lie in the same plane

Coplanar non-concurrent forces: All forces do not meet at a point, but lie in a single plane

Non-coplanar parallel forces: All the forces are parallel to each other, but not in same plane

Non-coplanar concurrent forces: All forces do not lie in the same plane, but their lines of action pass through a single point

Non-coplanar non-concurrent forces: All forces do not lie in the same plane and their lines of action do not pass through a single point

For ring:

MOI about diametric axis:

MOI about polar axis:

Linear mass density:

m=λ×2πR

n = 2

Without knowing the weight of the ladder, coefficient of friction between the ladder and floor cannot be found.

Ft=mat=mrα

Impulse = change in momentum

= mv₂ - mv₁

= 1 (10) - 1 (-25) = 35 N - s 

For a square of side a, the area moment of inertia of square with respect to one diagonal is

Here a = 1

When lift is at rest: T = mg

When lift is accelerating upward: T_U = mg + ma

When lift is accelerating downward: T_D = mg - ma

T_U = 2T_D

mg + ma = 2mg - 2ma

a = g/3

Moment of inertia of a rectangle about an axis though its center of gravity is bd³/12 and of a square is a⁴/12.

Mass distribution is minimum about centroidal axis, so moment of inertia of an area always least with respect to ​centroidal axis

An object is said to be in equilibrium when there is no external net force acting on it. When an object is in equilibrium, it does not accelerate. If it had a velocity, the velocity remains constant; if it was at rest, it remains at rest.

If all the forces acting on a particular object add up to zero and have no resultant force, then it’s in translational equilibrium. Examples would be a book resting on a bookshelf, or someone walking at a steady, constant speed.

An object that’s not rotating or doing so at a steady speed, the sum of the torques acting on it equaling zero, is at rotational equilibrium. Some examples of this are a Ferris wheel turning at a constant velocity, two children of equal weight balanced on either side of a seesaw, or the Earth rotating on its axis at a steady speed.

Total energy of the disc rolling down on an inclined plane is

Total rotational energy is

Velocity = mass × acceleration

∴ 50 = 10 × a

⇒ a = 5 m/sec²

Velocity after 2 seconds,

v = u + at

= 1 + 5 × 2

= 11 m/sec

If the number of independent static equilibrium equations is not sufficient for solving all the external and internal forces in a system, then the system is said to be statically indeterminate.

Degree of static indeterminacy = Total no. of unknowns (External and internal) – Number of independent equations of equilibrium.

Now in this problem,

No. of unknowns = 4 (RAH, RAV, M and RB)

No. of independent equations of equilibrium = 3 (∑H = 0, ∑V = 0, ∑M = 0)

∴ Degree of static indeterminacy - 4 - 3 = 1

Angular motion θ = 3t + t³

Motion covered in 2 seconds = 3 * 2 + 2³ = 14 rad. So the angular position after 2 seconds will be 14 rad.

at t = 5 sec, s = 450 m

at t = 10 sec

s = 450 + 700 = 1150

Equation (ii) – Equation (i), we get

a = 10 m/s²