Chapter 4, Problem 24P

One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-49 (a) If the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.25 m/s²by the upper cord, calculate the tension in each cord.

Figure 4-49
Problem 25

To determine

The tension in the cords hanging the buckets.

Answer to Problem 24P

Solution:

Tension in lower and upper strings are $31.36\text{N}$ and $62.72\text{ N}$ respectively.

Explanation of Solution

Given:

  $m=3.2\text{ kg}$ ............................................................................... (Mass of each bucket)

  $g=9.8{\text{ m/s}}^{\text{2}}$ ................................................................... (Acceleration due to gravity)

To find: (I) $T_1=?$ .................................................................................... (Tension in lower cord)

(II) $T_2=?$ .................................................................................... (Tension in upper cord)

Formula:

From Newton’s second law of motion:

  ${\displaystyle \sum F}=ma$

Where, m is mass and a is acceleration.

Calculation:

The buckets are at rest. This means that acceleration of both buckets is $a=0{\text{ m/s}}^{\text{2}}$ .

1. Fromfree body diagram of lower bucket.

  

  $\begin{array}{l}{\displaystyle \sum F=ma}\\ T_1 - mg=m\left(0\right)\\ T_1-\left(3.2\right)\left(9.8\right)=(3.2)(1.25)\\ T_1=35.36\text{N}\end{array}$

2. From free body diagram of upper bucket.

  

  $\begin{array}{l} {\displaystyle \sum F=ma}\\ T_2 -T_1 - mg=m\left(0\right)\\ T_2-31.36-\left(3.2\right)\left(9.8\right)=0\\ T_2=62.72\text{N}\end{array}$

Conclusion: Using Newton’s second law and free body diagram, the tension in the lower string as $31.36\text{N}$ and tension in upper string as $62.72\text{N}$.

To determine

The tension in the cords hanging the buckets when acceleration is $1.25{\text{m/s}}^{\text{2}}$ .

Answer to Problem 24P

Solution:

Tension in lower and upper strings are $35.36\text{N}$ and $70.72\text{N}$ respectively.

Explanation of Solution

Given:

  $m=3.2\text{ kg}$ ............................................................................... (Mass of each bucket)

  $g=9.8{\text{ m/s}}^{\text{2}}$ ................................................................... (Acceleration due to gravity)

  $a=1.25{\text{m/s}}^{\text{2}}$ .............................................................. (Acceleration of both buckets)

To find: (I) $T_1=?$ ................................................................................... (Tension in lower cord)

(II) $T_2=?$ .................................................................................. (Tension in upper cord)

Formula:

  ${\displaystyle \sum F=ma}$

Calculation:

1. From, Free body diagram of lower bucket.

  

  $\begin{array}{l}{\displaystyle \sum F=ma}\\ T_1 - mg=m\left(0\right)\\ T_1-\left(3.2\right)\left(9.8\right)=(3.2)(1.25)\\ T_1=35.36\text{N}\end{array}$

2. From, Free body diagram of upper bucket.

  

  $\begin{array}{l}{\displaystyle \sum F=ma}\\ T_2 -T_1 - mg=ma\\ T_2-35.36-\left(3.2\right)\left(9.8\right)=(3.2)(1.25)\\ T_2=70.72\text{N}\end{array}$

Conclusion: Using Newton’s second law and free body diagram, the tension in the lower string as $35.36\text{N}$ and tension in upper string as $70.72\text{N}$ .