Engineering Mechanics - 2 - Mechanical Engineering MCQ

Free body diagram of a ladder resting on a frictional surface is:

The force of friction will be upwards at its upper end and towards the wall at its lower end.

As the ground is smooth (but the wall does not), so f_A = 0 and f_B ≠ 0

So the force of friction will be upwards at its upper end.

Normal component: a_n = v²/r = (10)²/(25) = 4

The person exerts a downward force on the scale, and the scale pushes up on the person with an equal (but opposite) force, F_N. Thus, the scale reading is F_N, the magnitude of the normal force.

F_N – F_W = ma ⇒ F_N – mg = ma

⇒ F_N = ma + mg = m (a + g) = (W/g) (a + g) = (800/10) (10 + 5)

F_N = 1200 N

Resultant between two forces P and Q, acting at an angle θ is defined as:

A fan is claimed to increase the velocity of air to a specified value while consuming electric power at a specified rate. The electric power input will be equal to the rate of increase of the kinetic energy of air

Equilibrium is the status of the body when it is subjected to a system of forces. If the resultant force is equal to zero it implies that the net effect of the system of forces is zero this represents the state of equilibrium.

Using Newton’s Second law of motion:

R−mg=ma⇒R=m(g+a)R−mg=ma⇒R=m(g+a)

So apparent weight will be greater than the true weight.

Force = T - W

ma = T - W

Acceleration is given as

Tension is given as

T = mg + ma = m (g + a)

T = 1000 (10 + 2)

T = 12000 N

By taking the values of θ, given in the options we can conclude that for θ = 60⁰, T₂ is minimum.

Acccording to Newton's Second Law of Motion:

F=ma=100×5=500kg−m/s2

Conservation of energy:

P.E1+K.E1=P.E2+K.E2P.E1+K.E1=P.E2+K.E2

As there is no loss of velocity so

K.E=12mv2⇒K.E1=K.E2K.E=12mv2⇒K.E1=K.E2

P.E1=P.E1⇒h1=h2⇒h2=2m

Initially checking whether towing force can overcome the static friction force or not.

Friction force (f) = μN = μmg = 0.3 × 50 × 10 = 150 N

∴ Net force on box = 400 – 150 = 250 N

⇒ 250 = ma

Velocity after 5 second

V = u + at = 0 + 5 × 5 = 25 m/s

v_x = v_o = Horizontal component of velocity

v_y = Vertical component

Time taken by the body to cover vertical distance H

Let at a distance of x, weight to be pulled is 10 kg of bucket and x kg of rope weight.

So total weight is (10+x)

Kinetic energy  represents energy the mass possesses by virtue of its motion. Likewise, potential energy (PE=mgh) represents energy the mass possesses by virtue of its position.

E=KE+PE

E is the total energy of the mass: i.e., the sum of its kinetic and potential energies. It is clear that E is a conserved quantity: i.e., although the kinetic and potential energies of the mass vary as it falls, its total energy remains the same and one form of energy is being converted into other form of energy.

In principal axes of a section product of inertia is zero.

Product of inertia:- The moment of inertia between any two mutually perpendicular axis in the plane of area is the produce of inertia.

Product of inertia,

Ixy=∫dA.xy

The energy dissipation during impact is called by the term, coefficient of restitution, a scalar quantity

Position (x) = t³ – 2t² + 10t – 4

Velocity (v) =

At t = 3 s

V = 3 × 3² – 4 × 3 + 10 = 25 m/s

Rpm is just a measure of how many times a point is revolving per minute.

In a single revolution, a point rotates an angle = 2π

So, the total angle a point is rotating per minute = N x 2 π

The angular velocity is a measure of how fast it is rotating = Angle Rotated/Time Required = 2π N/60

Horizontal range,

For maximum horizontal range.

Sin 2α = 1

Or 2α = 90°

Or α = 45°