ω = 2πN/60 = 2π(240)/60 = 25.13 m/s and n = I/r = 1.25/0.3 = 4.167

**(a) Force on the piston rod, **i.e., piston effort(F_{p}) :

we know that net load onthe piston,

F_{L} = (P_{1} - P_{2}) x (π/4) x D^{2}

F_{L} = (1.125 x 10^{6} - 0.125 x 10^{6}) x (π/4) x 0.3^{2} = 70685.83 N

Inertia force on reciprocating parts is given by

F_{1} = m_{R }ω^{2}r (cos*θ + cos 2**θ/n*)

F_{1} = 60(25.13)^{2} x 0.3 x (cos 60 + cos2 x 60/4.167) = 4319.68 N

Therefore, Force on the piston rod or piston effort

F_{P} = F_{L} - F_{I} = 70685.83 - 4319.68 = 66.366 kN.

**(b) Turning moment on the crankshaft (T): **

We know that turning moment on the crankshaft,