Question

A slider crank mechanism is shown in the figure. At some instant, the crank angle is ${45}^o$ and a force of 40 N is acting towards the left on the slider. The length of the crank is 30 mm and the connecting rod is 70 mm. Ignoring the effect of gravity, friction and inertial forces, the magnitude of the crankshaft toruqe(in Nm) needed to keep the mechanism in equlibrium is (correct to two decimal places).

Answer 1

$30sin{45}^o={70}^{o}sin\beta$
$sin\beta =\frac{30sin{45}^o}{70}=0.30345$
$\beta ={17.6406}^o$
$F_{c}cos\beta =40$
$F_c=\frac{40}{cos\beta }=41.9737N$
$F_t=F_{c}sin(\theta +\beta )$
$=41.9737\times sin({17.6406}^o+{45}^o)$
$F_t=37.2785N$
$T=F_t\times r=37.2785\times 0.030$
$=1.1183Nm=1.118Nm$
$=1.12Nm$