Question

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 34.5 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 kN?

(a) 2.238

(b) 2.508

(c) 2.754

(d) 2.908

I got this question in unit test.

I'm obligated to ask this question of Balancing of V-engines topic in section Balancing of Reciprocating Masses of Machine Dynamics

Answer 1

Right option is (b) 2.508

Explanation: Maximum resultant secondary force is given by the equation:

\(\sqrt{2}\) (m/n)xω^2.r

substituting the values we get

Fsmax = 2508 N.