Farm Power · Farm Power · 2 marks · Numerical answer
An engine crank radius is 250 mm and connecting rod length1000 mm. At100° from head dead centre, net piston force along motion is 35 kN. Find crank torque in kN m, to two decimals.
Rod obliquity satisfies $\sin\beta=(r/l)\sin\theta=.25\sin100^\circ$, hence β≈14.25°. Tangential crank force is $F\sin(\theta+\beta)/\cos\beta$. Torque is $35(.25)\sin(100^\circ+\beta)/\cos\beta=\mathbf{8.23}$ kN m. Rod obliquity must be included.
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