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# Machine Dynamics MCQ – Variation of Tractive Force

The resultant unbalanced force due to the two cylinders, along the line of stroke, is known as ______

a) Tractive force

b) Swaying couple

c) Hammer blow

d) Hammer couple

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Answer: a

Explanation: The resultant unbalanced force due to the two cylinders, along the line of stroke, is known as tractive force. The value of tractive force depends on the angle of the crank.

Tractive force is produced due to unbalanced forces perpendicular to the line of stroke of action.

a) True

b) False

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Answer: b

Explanation: Tractive force is produced due to unbalanced forces along the line of stroke of action. The value of tractive force depends on the angle of the crank.

At which of the following angles in degrees does the tractive force attains a maximum value?

a) 135

b) 45

c) 90

d) 60

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Answer: a

Explanation: The tractive force is maximum or minimum when (cos θ – sin θ) is maximum or minimum, differentiating it with respect to theta gives tan θ = -1

hence maximum/minimum value occurs at 135 or 315 degrees

At which of the following angles in degrees does the tractive force attains a minimum value?

a) 315

b) 45

c) 90

d) 60

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Answer: a

Explanation: The tractive force is minimum when (cos θ – sin θ) is minimum, differentiating it with respect to theta gives tan θ = -1

hence maximum/minimum value occurs at 135 or 315 degrees.

Find the maximum value of tractive force in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a) 6978

b) 7574

c) 6568

d) 7374

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Answer: a

Explanation: The maximum value of tractive force is given by

+√2 (1-c)mω^{2}r

substituting the values we get

F = 6978 N.

Find the variation in tractive force in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a) 13956

b) 17574

c) 16568

d) 17374

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Answer: a

Explanation: The variation in tractive force is given by

±√2 (1-c)mω^{2}r

substituting the values we get

F = 13956 N.

Find the minimum value of tractive force in newtons from the following data:

m = 100Kg

ω = 300 rpm

radius = 0.6m

c = 2/3

a) -6978

b) -7574

c) -6568

d) -7374

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View Answer**

Answer: a

Explanation: The minimum value of tractive force is given by

-√2 (1-c)mω^{2}r

substituting the values we get

F = 6978 N.

Tractive force attains a maxima or minima when the crank angle is 90 degrees.

a) True

b) False

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Answer: b

Explanation: The tractive force is maximum or minimum when the expression (cos θ – sin θ ) is maximum or minimum, differentiating it with respect to theta gives tan θ = -1

hence maximum/minimum value occurs at 135 or 315 degrees.