Chapter 11, Problem 11.1P

(a)

To determine

The relation among the six lengths.

Answer to Problem 11.1P

The relation between the lengths is $k_1\left(l_1-L_1\right)=k_2\left(l_2-L_2\right)=k_3\left(l_3-L_3\right)$.

Explanation of Solution

In the given case, $k_1$, $k_2$ and $k_3$ are the spring constants of the three springs, $m_1$ and $m_2$ are the masses of the two carts, $x_1$ and $x_2$ are the position of the two carts from its position of equilibrium.

The net force acting on each of the springs is same in equilibrium position and each of the springs experiences either tension of compression. The restoring force of the spring is directly proportional to the compressed length according to Hooke’s law.

    $F=-kx$        (I)

Here, $F$ is restoring force, $k$ is the spring constant and $x$ is the compressed length of each spring.

Rewrite equation (I) for the force acting on the first spring.

    $F_1=-k_1{x}_1$        (II)

Here, $k_1$ is the spring constant of the first spring and $x_1$ is the compressed length of the first spring.

The compressed length of the first spring can be written as

    $x_1=l_1-L_1$        (III)

Here, $l_1$ is the un-stretched length and $L_1$ is the equilibrium length of the first spring.

Substituting equation (III) in equation (II).

     $F_1=-k_1\left(l_1-L_1\right)$        (IV)

Similarly, the equation for the restoring force on the second spring can be written as

    $F_2=-k_2\left(l_2-L_2\right)$        (V)

Here, $l_2$ is the un-stretched length and $L_2$ is the equilibrium length of the second spring.

And the equation for the restoring force on the third springs is

    $F_3=-k_3\left(l_3-L_3\right)$        (VI)

Here, $l_3$ is the un-stretched length and $L_3$ is the equilibrium length of the second spring.

Conclusion:

The forcing acting on each of the spring is same at equilibrium. Hence equate equation (IV), (V) and (VI).

    $\begin{array}{c}-k_1\left(l_1-L_1\right)=-k_2\left(l_2-L_2\right)=-k_3\left(l_3-L_3\right)\\ k_1\left(l_1-L_1\right)=k_2\left(l_2-L_2\right)=k_3\left(l_3-L_3\right)\end{array}$

Therefore, the relation between the lengths is $k_1\left(l_1-L_1\right)=k_2\left(l_2-L_2\right)=k_3\left(l_3-L_3\right)$.

(b)

To determine

The validity of the equation.

Answer to Problem 11.1P

The relations are proved.

Explanation of Solution

Two different forces act on the first cart. The extensive force of the first spring and the compressive force of the second spring.

Write the equation for the extensive force on the first cart.

    $F_1=-k_1{x}_1$        (VI)

Here, $F_1$ is the extensive force, $k_1$ is the spring constant of the first spring and $x_1$ is the expanded length from equilibrium for the first spring.

Write the equation for the compressive force acting on the first cart by the second spring is,

    $F_2=k_2\left(x_2-x_1\right)$        (VII)

Here, $F_2$ is the compressive force, $k_2$ is the spring constant of the second spring and $x_2-x_1$ is the compressed distance of the second spring.

Write the equation for the net force acting on the first cart suing Newton’s second law of motion.

    $F_{net}=m_1{\dot{x}}_1$        (VIII)

Here, $F_{net}$ is the net force acting on the first cart, $m_1$ is the mass of the first cart, ${\dot{x}}_1$ is the acceleration of the first cart.

The net force acting on the first cart is the vector sum of the forces $F_1$ and $F_2$.

    $F_{net}=F_1+F_2$        (IX)

Substitute equation (VIII), (VII) and (VI) in equation (IX)

    $\begin{array}{c}{m}_1{\dot{x}}_1=\left(-k_1{x}_1\right)+\left(k_2\left(x_2-x_1\right)\right)\\ =-(k_1+k_2)x_1+k_2{x}_2\end{array}$        (X)

Consider the case of the case of the second cart. The forces acting on the second cart is the extensive force of the second spring and compressive force of the third spring.

Write the equation for the compressive acting on the second cart is,

    $F_3=-k_3{x}_2$        (XI)

Here, $F_3$ is the compressive force, $k_3$ is the force constant on the third spring and  $x_2$ is the compressed length of the third spring from its equilibrium.

Write the equation for the extensive force acting on the first cart by the second spring.

    $F_2=-k_2\left(x_2-x_1\right)$        (XII)

Here, $F_2$ is the extensive force, $k_2$ is the spring constant of the second spring, $x_2-x_1$ is the expanded distance of the second spring.

Write the equation for the net force acting on the second cart suing Newton’s second law of motion.

    $F_{net}=m_2{\dot{x}}_2$        (XIII)

Here, $F_{net}$ is the net force acting on the second cart, $m_2$ is the mass of the second cart, ${\dot{x}}_2$ is the acceleration of the second cart.

The net force acting on the second cart is

    $F_{net}=F_2-F_3$        (XIV)

The negative sign indicates that the force $F_3$ is compressive force or is in the opposite direction to the force $F_2$.

Substitute equation (XIII), (XII) and (XI) in equation (XIV).

      $\begin{array}{c}{m}_2{\dot{x}}_2=-k_2\left(x_2-x_1\right)-k_3{x}_2\\ =-k_2{x}_2+k_2{x}_1-k_3{x}_2\\ =k_2{x}_1-\left(k_2+k_3\right)x_2\end{array}$        (XV)

Conclusion:

Therefore, the net force on first cart is $m_1{\dot{x}}_1=\left(-k_1{x}_1\right)+\left(k_2\left(x_2-x_1\right)\right)$ and net force acting on the second cart is $m_2{\dot{x}}_2=k_2{x}_1-\left(k_2+k_3\right)x_2$. Hence, the relations are proved.