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).

The correct answer is 33.33.

1. Understanding the System Configuration
As shown in the provided schematic image, a machine of mass $m$ is supported on two vertical mounts (springs), each having a stiffness $k$. The machine is subjected to a dynamic vertical external force $F\left(t\right)$ directed upwards. Since the two mounts are connected in parallel to support the machine, the equivalent stiffness of the system, $k_{\text{eq}}$, is the sum of the stiffnesses of the individual mounts:
$k_{\text{eq}}=2k$
Given:
• Mass of the machine, $m=200\text{kg}$
• Stiffness of each mount, $k=10\text{kN/m}=10000\text{N/m}$
Therefore, the total equivalent stiffness is:
$k_{\text{eq}}=2\times 10000=20000\text{N/m}$

2. Natural Frequency of the System
The undamped natural frequency ${\omega }_n$ of the system is calculated as:
${\omega }_n=\sqrt{\frac{k_{\text{eq}}}{m}}$
Substituting the given values:
${\omega }_n=\sqrt{\frac{20000}{200}}=\sqrt{100}=10\text{rad/s}$

3. Frequency Ratio and Dynamic Parameters
The external harmonic force acting on the machine is given by:
$F\left(t\right)=50cos\left(5t\right)$
Comparing this with the standard harmonic force equation $F\left(t\right)=F_{0}cos\left(\omega t\right)$, we have:
• Amplitude of the external force, $F_0=50\text{N}$
• Excitation frequency, $\omega =5\text{rad/s}$
The frequency ratio $r$ is:
$r=\frac{\omega }{{\omega }_n}=\frac{5}{10}=0.5$

4. Transmissibility Ratio (TR)
Assuming no damping ($\xi =0$), the transmissibility ratio is given by:
$\text{TR}=\frac{1}{|1-r^2|}$
Since $r=0.5<1$:
$\text{TR}=\frac{1}{1-{0.5}^2}=\frac{1}{1-0.25}=\frac{1}{0.75}=\frac{4}{3}\approx 1.333$

5. Dynamic Force Transmitted to the Ground
The total dynamic force transmitted to the ground ($F_T$) is:
$F_T=\text{TR}\times F_0=1.333\times 50=66.67\text{N}$
Because the machine is symmetrically supported by two identical mounts in parallel, this total transmitted force is shared equally between the two mounts:
$F_{\text{each}}=\frac{F_T}{2}=\frac{66.67}{2}=33.33\text{N}$