Question

A slider crank mechanism with crank radius 200 mm and connecting rod length 800 mm is shown. The crank is rotating at 600 rpm in the counter clockwise direction. In the configuration shown, the crank makes an angle of ${90}^o$ with the sliding direction of the slider, and a force of 5 kN is acting on the slider. Neglecting the inertia forces, the turning moment on the crank (in kNm) is

Answer 1

Method I :
Turning moment on crankshaft,
$T=F_T\times r=\frac{F}{cos\beta }\times cos\beta \times r$
$=F\times r=5\times 0.2=$1 kNm

Method II:

$sin\beta =\frac{Perpendicular}{Hypotenuse}$
$sin\beta =\frac{200}{800}=\frac{1}{4}=0.25$
$\beta =si{n}^{-1}0.25={14.478}^o$
Turning moment on crankshaft,
$T=F_{t}r=\frac{F}{cos\beta }sin(\theta +\beta )\times r$
$\frac{5}{cos14.478}sin[{90}^o+{14.478}^o]\times 0.2$
$=1kNm$