# Machine Dynamics Questions and Answers – Primary and Secondary Unbalanced Forces of Reciprocating Masses

This set of Machine Dynamics Question Bank focuses on “Primary and Secondary Unbalanced Forces of Reciprocating Masses”.

1. The primary unbalanced force is maximum when the angle of inclination of the crank with the line of stroke is
a) 0°
b) 90°
c) 180°
d) 360°

Explanation: The primary unbalanced force is maximum, when θ = 0° or 180°. Thus, the primary force is maximum twice in one revolution of the crank. The maximum primary unbalanced force is given by
FP(max) = m⋅ω2⋅r

2. The partial balancing means
a) balancing partially the revolving masses
b) balancing partially the reciprocating masses
c) best balancing of engines
d) all of the mentioned

Explanation: To balance the reciprocating masses partially is known as partial balancing.

3. In order to facilitate the starting of locomotive in any position, the cranks of a locomotive, with two cylinders, are placed at ______________ to each other.
a) 45°
b) 90°
c) 120°
d) 180°

Explanation: Due to the partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke.

4. In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to
a) minimise the effect of primary forces
b) minimise the effect of secondary forces
c) have perfect balancing
d) start the locomotive quickly

Explanation: The secondary unbalanced force is maximum, when θ = 0°, 90°,180° and 360°. Thus, the secondary force is maximum four times in one revolution of the crank. Keeping large distance between connecting rod and crank, minimises the effect of secondary forces.

5. If c be the fraction of the reciprocating parts of mass m to be balanced per cyclinder of a steam locomotive with crank radius r, angular speed ω, distance between centre lines of two cylinders a, then the magnitude of the maximum swaying couple is given by
a) 1 – c / 2 mrω2a
b) 1 – c / √2mrω2a
c) √2(1 – c)mrω2a
d) none of the mentioned

Explanation: The swaying couple is maximum or minimum when (cosθ+sinθ) is maximum or minimum. For (cosθ+sinθ) to be maximum or minimum
∴ tanθ =1 or θ = 45° or 225°
Maximum value of the swaying couple = 1 – c / √2mrω2a

6. The swaying couple is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to
a) 45° and 135°
b) 90° and 135°
c) 135° and 225°
d) 45° and 225°

Explanation: The swaying couple is maximum or minimum when (cosθ+sinθ) is maximum or minimum. For (cosθ+sinθ) to be maximum or minimum
∴ tanθ =1 or θ = 45° or 225°

7. The tractive force is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to
a) 90° and 225°
b) 135° and 180°
c) 180° and 225°
d) 135° and 315°

Explanation: The tractive force is maximum or minimum when (cos θ – sin θ ) is maximum or minimum. For (cos θ – sin θ ) to be maximum or minimum,
∴ tanθ = −1 or θ =135° or 315°

8. The swaying couple is due to the
a) primary unbalanced force
b) secondary unbalanced force
c) two cylinders of locomotive
d) partial balancing

Explanation: The unbalanced forces along the line of stroke for the two cylinders constitute a couple. This couple has swaying effect about a vertical axis, and tends to sway the engine alternately in clockwise and anticlockwise directions. Hence the couple is known as swaying couple.

9. In a locomotive, the maximum magnitude of the unbalanced force along the perpendicular to the line of stroke, is known as
a) tractive force
b) swaying couple
c) hammer blow
d) none of the mentioned

Explanation: The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. The effect of hammer blow is to cause the variation in pressure between the wheel and the rail.

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