Chapter 3, Problem 64SP

A 300-g mass hangs at the end of a string. A second string hangs from the bottom of

that mass and supports a 900-g mass. (a) Find the tension in each string when the

masses are accelerating upward at $0.700{\text{ m/s}}^2$. Don’t forget gravity. (b) Find the

tension in each string when the acceleration is $0.700{\text{ m/s}}^2$ downward.

To determine

The tension in eachstring when the masses are accelerating in upward direction at $0.700{\text{ m/s}}^2$.

Answer to Problem 64SP

Solution:

$12.6\text{ N}$ and $9.45\text{}\text{ N}$

Explanation of Solution

Given data:

The mass of the block that is hanging on the first string is $300\text{ g}$.

The mass of the block that is hanging on the second string is $\text{900 g}$.

Upward acceleration of the masses is $0.700{\text{ m/s}}^2$.

Formula used:

From the Newton’s second law of the motion, the expression of the force is

${\displaystyle \sum F_y}=m{a}_y$

Here, ${\displaystyle \sum F_y}$ is the net force in y- direction, $m$ is the mass of block, and $a_y$ is acceleration of the block.

Explanation:

Draw the free body diagram of the system by analyzing the problem:

In the above diagram, $300\text{ g}$ is mass of the block that hangs on the first string, $T_1$ is the tension in the string, $W_1$ is the weight of the block that hangs on the string, $900\text{ g}$ is mass of the block that hangs on the second string which is attached by the first string, $m_1$ is the mass of the $300\text{ g}$ block, $T_2$ is the tension in the second string, $W_2$ is the weight of the block that hangs on the second string, and $m_2$ is the mass of the $900\text{ g}$ block.

Draw the free body diagram of the masses separately when acceleration is in upward direction:

Consider free body diagram of the $900\text{ g}$ block and recall the expression of Newton’s second law of the motion:

${\displaystyle \sum F_y}=m{a}_y$

Consider the direction of the upward forces as positive and the direction of the downward forces as negative. Therefore,

$T_2-m_{2}g=m_{2}a$

Here, $a$ is the acceleration of the block.

Substitute $\text{900 g}$ for $m_2$ and $0.700{\text{ m/s}}^2$ for $a$

$T_2-\left(900\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)=\left(900\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(0.700{\text{ m/s}}^2\right)$

Solve the above equation for $T_2$

$\begin{array}{c}{T}_2-8.82\text{ N}=0.63\\ T_2=9.45\text{ N}\end{array}$

Consider free body diagram of the $300\text{ g}$ block and recall the expression of Newton’s second law of the motion:

${\displaystyle \sum F_y}=m{a}_y$

Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Therefore,

$T_1-T_2-m_{1}g=m_{1}a$

Substitute $300\text{ g}$ for $m_1$, $9.45\text{ N}$ for $T_2$, and $0.700{\text{ m/s}}^2$ for $a$

$T_1-9.45\text{ N}-\left(300\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)=\left(300\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(0.700{\text{ m/s}}^2\right)$

Solve the above equation for $T_2$

$\begin{array}{c}{T}_1-12.39\text{N}=0.21\text{ N}\\ T_1=12.\text{6N}\end{array}$

Conclusion:

Therefore, the value of the tension in the strings by which $300\text{ g}$ and 900 g massesarehanged is $12.6\text{ N}$ and $9.45\text{}\text{ N}$, respectively, when both the masses accelerating up.

To determine

The tension in each string when the masses are accelerating in the downward direction at $0.700{\text{ m/s}}^2$.

Answer to Problem 64SP

Solution:

$\text{10}\text{.9 N}$ and $8.19\text{}\text{ N}$

Explanation of Solution

Given data:

The mass of the block that is hanging on the first string is $300\text{ g}$.

The mass of the block that is hangingon the second string is $\text{900 g}$.

Downward acceleration of the masses is $0.700{\text{ m/s}}^2$.

Formula used:

From the Newton’s second law of the motion, the expression of the force is

${\displaystyle \sum F_y}=m{a}_y$

Here, ${\displaystyle \sum F_y}$ is the sum of forces in y- direction, $m$ is the mass of string, and $a_y$ is acceleration.

Explanation:

Draw the free body diagram of the each masses separately when acceleration is in downward direction:

Consider the free body diagram of the $900\text{ g}$ mass and recall the expression of Newton’s second law of the motion:

${\displaystyle \sum F_y}=m{a}_y$

Consider the direction of the upward forces as positive and the direction of the downward forces as negative. Therefore,

$T_2-m_{2}g=-m_{2}a$

Substitute $\text{900 g}$ for $m_2$ and $0.700{\text{ m/s}}^2$ for $a$

$T_2-\left(900\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)=-\left(900\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(0.700{\text{ m/s}}^2\right)$

Solve the above equation for $T_2$

$\begin{array}{c}{T}_2-8.82\text{ N}=-0.63\\ T_2=8.19\text{ N}\end{array}$

Consider the free body diagram of the $300\text{ g}$ mass and recall the recall the expression of Newton’s second law of the motion:

${\displaystyle \sum F_y}=m{a}_y$

Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Therefore,

$T_1-T_2-m_{1}g=-m_{1}a$

Substitute $300\text{ g}$ for $m_1$, $9.45\text{ N}$ for $T_2$, and $0.700{\text{ m/s}}^2$ for $a$

$T_1-8.19\text{ N}-\left(300\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)=-\left(300\text{ g}\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\right)\left(0.700{\text{ m/s}}^2\right)$

Solve the above equation for $T_1$

$\begin{array}{c}{T}_1-11.13\text{N}=-0.21\text{ N}\\ T_1=10.9\text{N}\end{array}$

Conclusion:

Therefore, the value of the tension in the strings by which the $300\text{ g}$ and 900 g massesarehanged is $\text{10}\text{.9 N}$ and $8.19\text{}\text{ N}$, respectively, when both the masses accelerating downward.