Chapter 4, Problem 78GP

(a)

To determine

The minimum force $F$ is needed to lift the piano (mass $M$ ) using the pulley apparatus.

Answer to Problem 78GP

The minimum force $\frac{Mg}{2}$ is needed to lift the piano (mass $M$ ) using the pulley apparatus.

Explanation of Solution

Formula used:

Apply the Newton’s second law of motion to determine the minimum force. The formula for force is $F=ma$ , where, $F$ is the force, $m$ is the mass, $a$ is the acceleration.

Calculation:

The weight leads to gravitational force on the object, the force exerted on the object depends upon the mass of the object and the acceleration due to gravity.

  $F_G=mg$

The force leads to push or pull which is a vector, so the friction exerted on other by the object is,

  $F_{friction}={\mu }_k{F}_N$

Where, $F_N$ is the normal force which is the object exert on the other and kinetic friction coefficient is ${\mu }_k$ and static friction coefficient is ${\mu }_s$ .

Apply the Newton’s second law of motion, net force is equal to the product of mass $\left(m\right)$ and acceleration $\left(a\right)$ . $T_1$ and $T_2$ are tension at section of rope $F_{T1}$ , $F_{T2}$ respectively.

  $\begin{array}{c}\sum F_y=ma\\ \sum F_y=0\\ \sum F_y=T_4\\ \sum F_y=T_1+T_2\end{array}$

  $T_1=T_2$

Substitute the value,

  $\begin{array}{l}\sum F_y=2T_1\\ T_1=\frac{mg}{2}\\ T_1=\frac{Mg}{2}\end{array}$

Conclusion:

Therefore, the minimum force $\frac{Mg}{2}$ is needed to lift the piano (mass $M$ ) using the pulley apparatus.

(b)

To determine

The tension in each section of rope: $F_{T1}$ , $F_{T2}$ , $F_{T3}$ and $F_{T4}$ .

Answer to Problem 78GP

The tension at each section of rope is $T_1=T_2=\frac{1}{2}Mg$ , and tension at $T_3$ is $\frac{3}{2}Mg$ , and tension at $T_4$ is $Mg$ .

Explanation of Solution

Formula used:

Apply the Newton’s second law of motion to determine the tension on each section of rope. The formula for force is $F=ma$ , where, $F$ is the force, $m$ is the mass, $a$ is the acceleration.

Calculation:

Consider the pulleys frictionless and mass less. Apply the Newton s second law of motion,

  $\begin{array}{l}\sum F=ma\\ \sum F=0\end{array}$

Then,

  $\begin{array}{c}0=T_1+T_2-T_3+F\\ T_3=T_1+T_2+F\end{array}$$\left(1\right)$

Where, $T_1$ , $T_2$

  $T_3$ and $T_4$ are the tension at each section of rope.

The value of $T_2$ is $\frac{Mg}{2}$ , and $T_1=T_2$ then substitute the value in equation $\left(1\right)$ ,

  $\begin{array}{l}{T}_3=T_1+T_2+F\\ T_3=\frac{Mg}{2}+\frac{Mg}{2}+F\\ T_3=\frac{3}{2}Mg\end{array}$

Therefore, the tension at each section of rope is $T_1=T_2=\frac{1}{2}Mg$ , and tension at $T_3$ is $\frac{3}{2}Mg$ , and tension at $T_4$ is $Mg$ .