Chapter 13.1, Problem 13.22P

The motor applies a constant downward force $F=1050$ lb to the cable used to raise the 4000-lb elevator E shown in the figure. The counterweight has a weight of 2000 lb and the elevator starts from rest. After the elevator travels 20 ft. determine (a) the velocity of the elevator. (b) the velocity of the counterweight.

To determine

(a)

The velocity of the elevator.

Answer to Problem 13.22P

$v_E=4.633ft/s$

Explanation of Solution

Given information:

The constant downward force is equal to $F=1050lb.$

Weight of elevator E is equal to $4000lb.$

Counterweight is equal to $2000lb.$

Elevator travels $20ft.$

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

$2x_A+2x_B+x_C=C$

$C$ - Constant

$2\frac{d{x}_A}{dt}+2\frac{d{x}_B}{dt}+\frac{d{x}_C}{dt}=0$

The kinetic energy of a particle is defined as,

$T=\frac{1}{2}m{v}^2$

Principle of work and energy is defined as,

$U_{1\to 2}=T_2-T_1$

Above equation states “If a particle moves from A₁ to A₂ under an action of force F, the work of the force F is equal to the change in kinetic energy.”

Calculation:

Length of both cables is constant, therefore

For cable 1,

$y_C+2y_E=0$

Differentiate,

$v_C+2v_E=0$

For cable 2,

$y_E+y_W=0$

Differentiate,

$v_E+v_W=0$

For elevator E,

Apply principle of work and energy,

$\begin{array}{l}{U}_{1\to 2}=T_2-T_1\\ 2F\left(20ft\right)+T\left(20ft\right)-\left(4000lb\right)\left(20ft\right)=\frac{1}{2}\left(\frac{4000lb}{32.2ft/s^2}\right)v_E{}^2\\ 2\left(1050lb\right)\left(20ft\right)+T\left(20ft\right)-\left(4000lb\right)\left(20ft\right)=\frac{1}{2}\left(\frac{4000lb}{32.2ft/s^2}\right)v_E{}^2\end{array}$

For counter weight,

Apply principle of work and energy,

$\begin{array}{l}{U}_{1\to 2}=T_2-T_1\\ \left(2000lb\right)\left(20ft\right)-T\left(20ft\right)=\frac{1}{2}\left(\frac{2000lb}{32.2ft/s^2}\right)v_W{}^2\end{array}$

But we know that,

$v_W=-v_E$

Therefore,

$\left(2000lb\right)\left(20ft\right)-T\left(20ft\right)=\frac{1}{2}\left(\frac{2000lb}{32.2ft/s^2}\right){\left(-v_E\right)}^2$

Add both equations,

$2\left(1050lb\right)\left(20ft\right)+\left(2000lb\right)\left(20ft\right)-\left(4000lb\right)\left(20ft\right)=\frac{1}{2}{v}_E{}^2\left(\frac{4000lb}{32.2ft/s^2}+\frac{2000lb}{32.2ft/s^2}\right)$

Therefore,

$v_E=4.633ft/s$

Conclusion:

The velocity of elevator E is equal to $v_E=4.633ft/s.$

To determine

(b)

The velocity of the counter weight.

Answer to Problem 13.22P

$v_W=-4.633ft/s$

Explanation of Solution

Given information:

The constant downward force is equal to $F=1050lb.$

Weight of elevator E is equal to $4000lb.$

Counterweight is equal to $2000lb.$

Elevator travels $20ft.$

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

$2x_A+2x_B+x_C=C$

$C$ - Constant

$2\frac{d{x}_A}{dt}+2\frac{d{x}_B}{dt}+\frac{d{x}_C}{dt}=0$

The kinetic energy of a particle is defined as,

$T=\frac{1}{2}m{v}^2$

Principle of work and energy is defined as,

$U_{1\to 2}=T_2-T_1$

Above equation states “If a particle moves from A₁ to A₂ under an action of force F, the work of the force F is equal to the change in kinetic energy.”

Calculation:

According to sub part a,

We have found,

$v_E+v_W=0$

And,

$v_E=4.633ft/s$

Therefore, the velocity of counter weight is equal to,

$v_W=-4.633ft/s$

Conclusion:

The velocity of counter weight is equal to $v_W=-4.633ft/s$