Question

When the engine is running at 750 rpm, the magnitude of primary forces is observed to be 500 N. If the crank and connecting rod are of lengths, 250 mm and 1000 mm, respectively, the magnitude of secondary force would be
1. 125 N
2. 250 N
3. 350 N
4. 450 N

Answer 1

Correct Answer - Option 1 : 125 N

Concept:

The unbalanced force due to reciprocating mass is given by

\({F_{un}} = mr{ω ^2}\left( {\cos \theta + \frac{{\cos 2\theta }}{n}} \right)\)

The direction of this force is constant in line but the magnitude changes continuously,

m × r × ω2 cosθ is called primary force and \(mr{ω ^2}\frac{{\cos 2\theta }}{n}\) is called secondary force.

For large values of n secondary force is much lesser than the primary force.

The maximum value of primary unbalanced force will be m × r × ω2 and the maximum value of secondary unbalanced force will be \(\frac{{mr{ω ^2}}}{n}\)

Calculation:

Given ω = 750 rpm, F_P = 500 N, r = 250 mm, l = 1000 mm,

In the given question, the crank angle is not given, so lets assume maximum primary force.

From the given, F_P = m × r × ω2 = 500 N,

⇒ n = l/r = 1000/250 = 4,

⇒ F_S = F_P/n = 500/4 = 125 N