--- title: "Two-mass system X2/F by free body diagram, F-V analogy" subject: "Control Systems" chapter: 2 key: "semester_3/control_systems:fbd-fv-m1-m2-k1-b1" source: https://www.bctengineer.com/q/semester_3/control_systems/fbd-fv-m1-m2-k1-b1 site: BCT Engineering --- # Two-mass system X2/F by free body diagram, F-V analogy Control Systems · Chapter 2: Component Modeling - Asked in: 2081 Bhadra, 6+2 marks - Video: [Translational Mechanical Systems (Solved Example)](https://www.youtube.com/watch?v=Av86LcNuiMQ) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/fbd-fv-m1-m2-k1-b1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2 ## Question Find the transfer function X2(s)/F(s) of the mechanical system shown below by constructing the free body diagram and writing necessary mathematical equations. Also draw F-V analogy circuit. [Figure: force F applied to mass M1 (displacement x1, friction fc1 with ground); M1 connected to mass M2 (displacement x2, friction fc2 with ground) by two parallel paths: spring K1 in series with damper B1 (junction displacement x3), and spring K2 directly; M2 connected to the wall through damper B2] ## Short answer Eliminate junction x3 using series impedance K1B1s/(K1+B1s) in parallel with K2 to get coupling Zc. Then X2(s)/F(s) = Zc/[ab+Zc(a+b)] where a = M1s^2+fc1s, b=M2s^2+(fc2+B2)s; denominator has a pole at s=0. F-V analogy: F→e, M→L, friction→R, spring→1/C; the K1-B1 series branch becomes parallel C(1/K1)||R(B1) in the shared mesh. ## Answer Displacements: $x_1$ ($M_1$), $x_2$ ($M_2$), and $x_3$ = junction of $K_1$ (on the $M_1$ side) and $B_1$ (on the $M_2$ side). $K_2$ connects the masses directly. #### Free body equations - $M_1$: $F$ applied; opposed by $M_1\ddot x_1$, $f_{c1}\dot x_1$, $K_1(x_1 - x_3)$, $K_2(x_1 - x_2)$. - Junction (massless): $K_1(x_3 - x_1) + B_1(\dot x_3 - \dot x_2) = 0$. - $M_2$: opposed by $M_2\ddot x_2$, $(f_{c2} + B_2)\dot x_2$, $B_1(\dot x_2 - \dot x_3)$, $K_2(x_2 - x_1)$. $$ \begin{aligned} M_1\ddot x_1 + f_{c1}\dot x_1 + K_1(x_1 - x_3) + K_2(x_1 - x_2) &= F \\ K_1(x_3 - x_1) + B_1(\dot x_3 - \dot x_2) &= 0 \\ M_2\ddot x_2 + (f_{c2} + B_2)\dot x_2 + B_1(\dot x_2 - \dot x_3) + K_2(x_2 - x_1) &= 0 \end{aligned} $$ #### Eliminating $x_3$ The series $K_1$–$B_1$ branch acts as $K_{eq} = \dfrac{K_1B_1s}{K_1 + B_1s}$, in parallel with $K_2$. Total coupling: $$ Z_c = K_2 + \frac{K_1B_1s}{K_1 + B_1s} = \frac{K_1K_2 + B_1(K_1 + K_2)s}{K_1 + B_1s} $$ With $a = M_1s^2 + f_{c1}s$ and $b = M_2s^2 + (f_{c2} + B_2)s$: $$ \begin{aligned} (a + Z_c)X_1 - Z_cX_2 &= F \\ -Z_cX_1 + (b + Z_c)X_2 &= 0 \end{aligned} $$ $$ \frac{X_2}{F} = \frac{Z_c}{ab + Z_c(a + b)} $$ #### Transfer function Multiplying by $(K_1 + B_1s)$ and writing $P = K_1K_2 + B_1(K_1 + K_2)s$: $$ \frac{X_2(s)}{F(s)} = \frac{K_1K_2 + B_1(K_1 + K_2)s}{s\left\{(K_1 + B_1s)\,s\,(M_1s + f_{c1})(M_2s + f_{c2} + B_2) + P\left[(M_1 + M_2)s + f_{c1} + f_{c2} + B_2\right]\right\}} $$ The pole at $s = 0$ appears because neither mass is tied to the wall by a spring. #### F-V analogy $F\to e$, $M\to L$, $B, f_c\to R$, $K\to 1/C$. Mechanical elements in parallel (same velocity difference) become series elements, and elements in series (same force) become parallel. So the shared branch between mesh 1 ($M_1$) and mesh 2 ($M_2$) is $C = 1/K_2$ in series with the parallel pair $1/K_1 \parallel B_1$; the small loop inside that pair is mesh 3 (junction $x_3$). ```text L1=M1 R=fc1 L2=M2 R=fc2+B2 +-LLLL---/\/\--+--------LLLL---/\/\--+ | | | (~) e C=1/K2 | | i1 | i2 | | +---+---+ | | C=1/K1 R=B1 | | | i3 | | | +---+---+ | | | | +--------------+---------------------+ ``` Mesh equations ($R_{c1} = f_{c1}$, $R_{c2} = f_{c2}$, $R_{B1} = B_1$, $R_{B2} = B_2$, $C_1 = 1/K_1$, $C_2 = 1/K_2$): $$ \begin{aligned} L_1\frac{di_1}{dt} + R_{c1}i_1 + \frac{1}{C_2}\int(i_1 - i_2)\,dt + \frac{1}{C_1}\int(i_1 - i_3)\,dt &= e \\ \frac{1}{C_1}\int(i_3 - i_1)\,dt + R_{B1}(i_3 - i_2) &= 0 \\ L_2\frac{di_2}{dt} + (R_{c2} + R_{B2})i_2 + \frac{1}{C_2}\int(i_2 - i_1)\,dt + R_{B1}(i_2 - i_3) &= 0 \end{aligned} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the junction x3 equation set to zero? A. The junction is massless, so forces from K1 and B1 balance B. x3 is fixed C. No damping exists there D. F acts at it Answer: A. A massless point has no inertia, so K1(x3 - x1) + B1(x3' - x2') = 0. (Key point: Junction equation) 2. K1 in series with B1 acts as what equivalent coupling? A. K1 + B1 s B. K1 B1 s / (K1 + B1 s) C. K1 B1 D. (K1 + B1 s)/K1 B1 s Answer: B. Series elements share the same force, so they combine like K1 and B1 s in series. (Key point: Series branch) 3. Why is there a pole at s = 0? A. Neither mass is tied to the wall by a spring B. F is a step C. M1 is zero D. Damping is zero Answer: A. The system can drift at constant velocity because nothing provides a static restoring force. (Key point: Pole at origin) 4. In the F-V analogy, how do mechanical series elements appear? A. As parallel electrical elements B. As series elements C. As mesh sources D. They disappear Answer: A. Series mechanical elements share force, so they become parallel (same-voltage) elements. 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