Question

A 6kg bucket of water is being pulled straight up by a string at a constant speed. I determined that the tension on the string was F = ma F = (6kg * 9.8 m/s2) * 0a F = 58.8 N Now its asking At a certain point the speed of the bucket begins to change. The bucket now has an upward constant acceleration of magnitude 3 m/s2. What is the tension in the rope now? The correct answer was "about 78N"

then Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude3

m/s2.

Now what is the tension in the rope?

Answer 1

Concepts and reason

The concept used to solve this problem is Newton’s second law of motion.

Calculate the tension force using the Newton’s second law. The net force on the bucket is the difference of tension force pulling the bucket upwards and weight of the bucket pulling the bucket down towards the ground. Rearrange the Newton’s second law equation to calculate the tension force.

Fundamentals

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The equation of the Newton’s second law is,

$\sum {\vec F = m\vec a}$

Here, $\sum {\vec F}$ is the net force on the object, $m$ is mass of the object, and $\vec a$ is the acceleration of the object.

The weight of a body is the force of gravitation of the earth that is,

$\vec w = m\vec g$

Here, $w$ is the weight of the body, $m$ is the mass, and $g$ is the acceleration due to gravity.

Take all the forces in upward direction as positive and all the force in downward direction as negative.

Use the Newton’s second law equation.

$\sum {\vec F = m\vec a}$

Two forces act on the bucket, the tension force, $\vec T$ in the upward direction so positive and weight, $\vec w$ in the downward direction so negative.

Substitute $\vec T - \vec w$ for $\sum {\vec F}$ in the above equation$\sum {\vec F = m\vec a}$.

$\vec T - \vec w = m\vec a$

Use the weight equation $\vec w = m\vec g$ and substitute $m\vec g$ for $\vec w$ in the above equation $\vec T - \vec w = m\vec a$ and rearrange to solve for$\vec T$.

$\begin{array}{c}\\\vec T - m\vec g = m\vec a\\\\\vec T = m\vec g + m\vec a\\\\\vec T = m\left( {\vec g + \vec a} \right)\\\end{array}$

The acceleration of the bucket is zero when the bucket is rising at constant speed.

The expression for the tension in terms of acceleration $\vec a$ and the acceleration due to gravity $\vec g$ is,

$\vec T = m\left( {\vec g + \vec a} \right)$

Substitute $6{\rm{ kg}}$ for$m$, $10{\rm{ m/}}{{\rm{s}}^2}$ for$\vec g$, and $0{\rm{ m/}}{{\rm{s}}^2}$ for $\vec a$ in the above equation $\begin{array}{c}\\\vec T = \left( {6{\rm{ kg}}} \right)\left( {10{\rm{ m/}}{{\rm{s}}^2} + 0{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 60{\rm{ N}}\\\end{array}$

The acceleration of the bucket is upward so it is positive.

Substitute $6{\rm{ kg}}$ for$m$, $10{\rm{ m/}}{{\rm{s}}^2}$ for$\vec g$, and ${\rm{3 m/}}{{\rm{s}}^2}$ for $\vec a$ in equation$\vec T = m\left( {\vec g + \vec a} \right)$. $\begin{array}{c}\\\vec T = \left( {6{\rm{ kg}}} \right)\left( {10{\rm{ m/}}{{\rm{s}}^2} + 3{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 78{\rm{ N}}\\\end{array}$

The acceleration of the bucket is directed in downward. So, it has negative sign. That is$- \vec a$. Thus, the tension in the rope is,

$\vec T = m\left( {\vec g - \vec a} \right)$

Substitute $6{\rm{ kg}}$ for$m$, $10{\rm{ m/}}{{\rm{s}}^2}$ for$\vec g$, and ${\rm{3 m/}}{{\rm{s}}^2}$ for$\vec a$.

$\begin{array}{c}\\\vec T = \left( {6{\rm{ kg}}} \right)\left( {10{\rm{ m/}}{{\rm{s}}^2} - 3{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = \left( {6{\rm{ kg}}} \right)\left( {7{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 42{\rm{ N}}\\\end{array}$

Ans:

The tension in the rope is$60{\rm{ N}}$.