GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 52 of 65

Marks: +1, -0

A slider crank mechanism is shown in the figure. At some instant, the crank angle is

45°

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 torque (in

{Nm}

) needed to keep the mechanism in equilibrium is ______________ (correct to two decimal places).

Solution:

30 \sin 45^{∘} \;\; =70 \sin β

\sin β \;\; =\;{30 \sin 45}/{70}=0.303045

β \;\; =17.6406^{∘}

F_{c} \cos β \;\; =40

F_{C} \;\; =\;{40}/{\cos β}=41.9737 {N}

F_{ {t}} \;\; =F_{c} \sin (θ+β)

\;\; =41.9737 × \sin (17.6406^{∘}+45^{∘})

F_{ {t}} \;\; =37.2785 {N}

T \;\; =F_{t} × r=47.2785 × 0.030

\;\; =1.1183 {N}- {m}=1.118 {N}- {m}

\;\; =1.12 {N}- {m}

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