Question

1. For the mechanical system illustrated below where a cart with a mass of $m = 1$ kg is sliding on the ground connected to the wall through a dashpot with a damping coefficient of $c =$ 10 N/mm/sec. dashpot $v(t)$ (Velocity) C M $f(t)$ (force) a) Write down the equation of motion and derive the transfer function between force and velocity of the mass, $G(s) = \frac{V(s)}{F(s)}$. b) What is the time constant of this system? c) What is the final velocity when a 1 N step force command is applied, $F(s) = \frac{1}{s}$.

          1. For the mechanical system illustrated below where a cart with a mass of $m = 1$ kg is sliding
on the ground connected to the wall through a dashpot with a damping coefficient of $c =$
10 N/mm/sec.
dashpot
$v(t)$ (Velocity)
C
M
$f(t)$ (force)
a) Write down the equation of motion and derive the transfer function between force and
velocity of the mass, $G(s) = \frac{V(s)}{F(s)}$.
b) What is the time constant of this system?
c) What is the final velocity when a 1 N step force command is applied, $F(s) = \frac{1}{s}$.
        
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1. For the mechanical system illustrated below where a cart with a mass of m = 1 kg is sliding
on the ground connected to the wall through a dashpot with a damping coefficient of c =
10 N/mm/sec.
dashpot
v(t) (Velocity)
C
M
f(t) (force)
a) Write down the equation of motion and derive the transfer function between force and
velocity of the mass, G(s) = (V(s))/(F(s)).
b) What is the time constant of this system?
c) What is the final velocity when a 1 N step force command is applied, F(s) = (1)/(s).

Added by Carla E.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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For the mechanical system illustrated below, where a cart with a mass of m = 1 kg is sliding on the ground connected to the wall through a dashpot with a damping coefficient of c = 10 N/mm/sec. a) Write down the equation of motion and derive the transfer function between force and velocity, F(s). b) What is the time constant of this system?
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Transcript

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00:01 And this lets the equation applying newton's second law that is if mass is equals to mx so we can write over here m plus f v2 so that is plus k1 plus k2 minus f v2 x2 minus k2 x2 is equal to zero now given values m1 is equal to 1 k2 k1 is equal to 4 k2 so x1 plus 6 x1 plus 9 x1 minus 3 x2 minus 5 x2 is equal to 0 now applying laplace transform so we would have 3x plus b so that is a x plus m b so x is equal to s square x s minus minus x zero so they would have 0 lx that the question xs assume there is no initial conditions x1 is equal to zero so s square…
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