Chapter 3, Problem 57SP

A 20-kg crate hangs at the end of a long rope. Find its acceleration (magnitude and

direction) when the tension in the rope is (a) 250 N, (b) 150 N, (c) zero, (d) 196 N.

To determine

The magnitude of the acceleration on thecrate when $20\text{ kg}$ mass of the crate hangs at the end of a long rope and tension in the rope is $250\text{ N}$.

Answer to Problem 57SP

Solution:

$2.7{\text{ m/s}}^2$ upward

Explanation of Solution

Given data:

The mass of the crate is $20\text{ kg}$.

The tension force in the rope is $250\text{ N}$.

Formula used:

FromNewton’s second law of motion, the expression of the sum of the force acting upon the crateis as follows:

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

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

Explanation:

Draw the free body diagram of the crate:

Here, $F_t$ is a tension force, $m$ is the mass of the crate, $a$ is the acceleration of the crate, and $g$ is acceleration due to gravity.

Recall the expression for Newton’s second law of motion when the motion of the crate is along the vertical direction:

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

Considering the direction of the forces, upward is taken aspositive and downward direction of the forces is taken as negative. Hence,

$F_t-mg=ma$

Rearrange the value for acceleration, a, of the crate:

$a=\frac{F_t-mg}{m}$

Substitute $250\text{ N}$ for $F_t$, $20\text{ kg}$ for $m$, and $\text{9}{\text{.80 m/s}}^2$ for $g$

$\begin{array}{c}a=\frac{\left(250\text{ N}\right)-\left(20\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)}{\left(20\text{ kg}\right)}\\ =\frac{53.8\text{ N}}{20\text{ kg}}\\ =2.7{\text{ m/s}}^2\end{array}$

Conclusion:

Therefore, the acceleration on the crate is $2.7{\text{ m/s}}^2$ along the upward direction.

To determine

The magnitude of the acceleration on the crate when $20\text{ kg}$ mass of the crate hangs at the end of a long rope and tension in the rope is $150\text{ N}$.

Answer to Problem 57SP

Solution:

$2.3{\text{ m/s}}^2$ downward

Explanation of Solution

Given data:

Mass of the crate that is hanging is $20\text{ kg}$.

Tension force is $150\text{ N}$.

Formula used:

From Newton’s second law of motion, the expression of the sum of the force acting upon the crateis as follows:

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

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

Explanation:

Draw the free body diagram of the crate:

Here, $F_t$ is a tension force, $m$ is the mass of the crate, $a$ is the acceleration of the crate, and $g$ is acceleration due to gravity.

Recall the expression for Newton’s second law of motion:

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

Considering the direction of the forces, upward is taken as positive and downward direction of the forces is taken as negative. Hence,

$F_t-mg=ma$

Rearrange the value for acceleration a:

$a=\frac{F_t-mg}{m}$

Substitute $150\text{ N}$ for $F_t$, $20\text{ kg}$ for $m$, and $\text{9}{\text{.80 m/s}}^2$ for $g$

$\begin{array}{c}a=\frac{\left(150\text{ N }\right)-\left(20\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)}{\left(20\text{ kg}\right)}\\ =\frac{\left(-46\text{ N}\right)}{\left(20\text{ kg}\right)}\\ =-2.3{\text{ m/s}}^2\end{array}$

Here, the value of $a$ is negative. Therefore, the free body diagram of the rope is as follows:

Conclusion:

The acceleration on the crate is $-2.3{\text{ m/s}}^2$ where the negative sign denotes that the direction of the acceleration is downward.

To determine

The magnitude of the acceleration on the crate when $20\text{ kg}$ mass of the crate hangs at the end of a long rope and tension in the rope is $\text{0 N}$ .

Answer to Problem 57SP

Solution:

$\text{9}{\text{.80 m/s}}^2$ downward

Explanation of Solution

Given data:

Mass of the crateis $20\text{ kg}$.

Tension force in the rope is $0\text{ N}$.

Formula used:

From Newton’s second law of motion, the expression of the sum of the force acting upon the rope is as follows:

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

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

Explanation:

Draw the free body diagram of the crate:

Here, $F_t$ is a tension force, $m$ is the mass of the crate, $a$ is the acceleration of the crate, and $g$ is acceleration due to gravity.

Recall the expression for Newton’s second law of motion:

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

Considering the direction of the forces, upward is taken as positive and downward direction of the forces is taken as negative. Hence,

$F_t-mg=ma$

Rearrange the value for acceleration a:

$a=\frac{F_t-mg}{m}$

Substitute $0\text{ N}$ for $F_t$, $20\text{ kg}$ for $m$, and $\text{9}{\text{.80 m/s}}^2$ for $g$

$\begin{array}{c}a=\frac{\left(0\right)-\left(20\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)}{\left(20\text{ kg}\right)}\\ =-9.80{\text{ m/s}}^2\end{array}$

The negative value of $a$ shows that its direction mentioned in the diagram should be along the downward direction.

Conclusion:

The acceleration on the crate is $\text{9}{\text{.80 m/s}}^2$ and the direction of the acceleration is downward.

To determine

The magnitude of the acceleration on the crate when $20\text{ kg}$ mass of the crate hangs at the end of a long rope and tension in the rope is $196\text{ N}$.

Answer to Problem 57SP

Solution:

$\text{0}{\text{.0 m/s}}^2$

Explanation of Solution

Given data:

The mass of the crateis $20\text{ kg}$.

The tension force in the rope is $196\text{ N}$.

Formula used:

From Newton’s second law of motion, the expression of the sum of the force acting upon the rope is as follows:

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

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

Explanation:

Draw the free body diagram of the crate:

Here, $F_t$ is a tension force, $m$ is the mass of the crate, $a$ is the acceleration of the crate, and $g$ is acceleration due to gravity.

Recall the expression for Newton’s second law of motion:

${\displaystyle \sum F_Y}=m{a}_Y$

Considering the direction of the forces, upward is taken as positive and downward direction of the forces is taken as negative. Hence,

$F_t-mg=ma$

Rearrange the value for acceleration of the crate:

$a=\frac{F_t-mg}{m}$

Substitute $196\text{ N}$ for $F_t$, $20\text{ kg}$ for $m$, and $\text{9}{\text{.80 m/s}}^2$ for $g$

$\begin{array}{c}a=\frac{\left(196\text{N }\right)-\left(20\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)}{\left(20\text{ kg}\right)}\\ =0{\text{ m/s}}^2\end{array}$

Conclusion:

The acceleration on the crate is ${\text{0 m/s}}^2$ when the tension force in the rope is $196\text{ N}$.