+1 vote
in Machine Dynamics by (34.3k points)
A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 23 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons?

(a) 7172

(b) 1672

(c) 1122

(d) 1272

This question was posed to me in an internship interview.

I need to ask this question from Balancing of V-engines topic in portion Balancing of Reciprocating Masses of Machine Dynamics

1 Answer

0 votes
by (38.6k points)
Right option is (b) 1672

For explanation I would say: Maximum resultant secondary force is given by the equation:

\(\sqrt{2}\) (m/n)xω^2.r

substituting the values we get

Fsmax = 1672 N.

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