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12ME2201ENGINEERING MECHANICS - Solutions

Part - A

1. Define resultant forceAny number of forces can be replaced into a single equivalent force called resultant force

which produces the same effect as that of all force.

2. State Principle of Transmissibility.The state of rest or motion of a rigid body is unaltered if a force acting on the body is

replaced by another force of the same magnitude and direction but acting anywhere on

the body along the line of action of the replaced force.

3. List out various types of Loads and supports.

Loads: Point load, uniformly distributed load and uniformly varying load

Supports: Pin / Hinged support, roller support and fixed support.

4. State Law of Coulomb friction.

The direction of friction force is always acts in a direction opposite to which the body

tends to move

The frictional force developed is independent of the area of contact

The total frictional force depends on the roughness of contact surfaces

The coefficient of kinetic friction is less than the co-efficient of static friction.

5. Locate the centroid of a semi circular area

0, 4r/3

6. State perpendicular axis theorem

The moment of inertia about ZZ axis which is perpendicular to XX and YY is the sum of

moment of inertia of XX axis and YY axis.

Izz = Ixx + Iyy

7. Write down the equation of motion of a particle

Fx = max

Fy = may

Fz = maz

8. State work energy principle

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The work of a particle from position 1 to 2 is change in its kinetic energies

U1-2 = KE2KE1

9 Define general plane motion of a rigid body

A motion combined with translation and rotation

10. Define motion about a fixed point

When a body moves in such a manner that one point, O remains fixed in space it can be

shown that the resulting displacement is a rotation about some axis through O with

angular velocity .

PartB (5 x 16 = 80)

11. a)

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b)

Fx= 8cos 30 -3-10cos60+6cos60 = 1.6N`Fy= 8sin 30 +5 -10sin60-6sin60 = -4.79N

R = Fx2 + Fy2 = 5.15N = 68

12. a)

b)

tan = P/dm

= 3.39

P = Wtan(+ )

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= 540 ( tan ( 3.64+ 2.86) = 61.53 N

The effort applied at the end of 300mm level may be found

P x 300 = 61.53 x62.5/2

= 6.41N

The efficiency of jack = 56%

Since the efficiency of the screw jack is more than 50% it is not self locking.

13. a)

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b)

14. a)

b)

Fx = max

N-mgcos 15 =0

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N= mg cos 15

Fy= may

175mg sin 15-k N = ma

175-mgsin 15 -0.3 mgcos 15 = 25a

A= 1.618m/s2.

15. a)

1) Angular velocity after 3 seconds =26rad/s2) Number of rotations in 3 seconds = 2403) Angular velocity after rotating twice= 7 rad/s4) Acceleration and velocity of A, at t=0. is 3m/s2. 12m/s

b)

Velocity =0.314m/s

Acceleration = 0.16m/s2

, 0.49m/s2.

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