Question

A machine of mass m =200 kg is supported on two mounts, each of stiffness k =10 kN/m. The machine is subjected to an external force (in N) F(t)=50 cos 5t. Assuming only vertical translatory motion, the magnitude of the dynamic force (in N) transmitted from each mount to the ground is (correct to two decimal places).

Answer 1

m = 200 kg, k = 10 kN/m = 10000 N/m
$k_{eq}=k+k=10000+10000=20000$ N/m
$F_t=50cos5t$
$F_0=50$ N
$\omega =5$ rad/s
${\omega }_n=\sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{20000}{200}}=\sqrt{100}$
$=10$ rad/s
$\frac{\omega }{{\omega }_n}=\frac{5}{10}=\frac{1}{2}$
No damping $\Rightarrow c=0\Rightarrow \xi =0$

Transmissibility, $ϵ=\frac{\sqrt{1+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2}}{\sqrt{{\{1-\left(\frac{\omega }{{\omega }_n}\right)\}}^2+{\left\{\frac{2\xi \omega }{{\omega }_n}\right\}}^2}}$
$(\xi =0)$
$=\frac{1}{\sqrt{{\{1-{\left(\frac{\omega }{{\omega }_n}\right)}^2\}}^2}}=\frac{1}{\sqrt{{\{1-{\left(\frac{1}{2}\right)}^2\}}^2}}$
$=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\left(\frac{4}{3}\right)$
$ϵ=\frac{F_t}{F_0}$
$\frac{4}{3}=\frac{F_t}{50}$
$F_t=\frac{200}{3}=66.666$ N
Force from each mount
$=\frac{F_t}{2}=\frac{66.666}{2}=33.333$ N