Question

A load of bricks with mass = 15.0 hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass = 27.6 is suspended from the other end of the rope, as shown in the figure. The system is released from rest.
1)What is the magnitude of the upward acceleration of the load of bricks?
2)What is the tension in the rope while the load is moving?

Answer 1

Concepts and reason

The concept required to solve this question is Newton’s second law of motion.

First, draw the free body diagram of all the force acting on the mass pulley system.

Later, calculate the magnitude of the acceleration by rearranging the newton’s second law.

Finally, calculate the tension by using the rearranged Newton’s second law equation.

Fundamentals

Acceleration is the rate of change of velocity.

The tension T in the rope is given by following expression:

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

1)

The following given figure is the free body diagram of a pully- mass system.

Here, y is the displacement of the masses and g is the acceleration due to gravity.

The direction of motion of the pulley – mass system depends on the masses hung on both the sides. As the masses are attached to both the ends of the rope, a tension T is generated.

Find the acceleration of the lighter mass m.

Let the acceleration of mass m be labeled as a and the acceleration of mass M as -a.

Use Newton’s second law of motion, to write the following equations:

$\begin{array}{c}\\T - Mg = - Ma\\\\T - mg = ma\\\end{array}$

Subtract first equation from second for acceleration as follows:

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

The acceleration is given by,

$a = \frac{{\left( {M - m} \right)g}}{{\left( {M + m} \right)}}$

Substitute $\left( {27.6{\rm{ Kg}}} \right)$ for M, $\left( {{\rm{15}}{\rm{.0 Kg}}} \right)$ for m and $\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)$ for g as follows:

$\begin{array}{c}\\a = \frac{{\left( {M - m} \right)g}}{{\left( {M + m} \right)}}\\\\ = \frac{{\left( {27.6{\rm{ Kg}} - 15{\rm{ Kg}}} \right)\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}{{\left( {27.6{\rm{ Kg}} + 15{\rm{ Kg}}} \right)}}\\\\ = 2.91{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}\\\end{array}$

As the load of bricks is less than that of the counterweight, so the acceleration of the bricks is in upward direction.

2)

Here, it is assumed that the pulley is frictionless, therefore the tension in the rope will be the same at every point on it.

The tension T in the rope is given as follows:

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

Substitute $\left( {9.81{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)$ for g, $\left( {{\rm{2}}{\rm{.91 m}} \cdot {{\rm{s}}^{ - 2}}} \right)$ for a and $\left( {{\rm{15}}{\rm{.0 Kg}}} \right)$ for mass m as follows:

$\begin{array}{c}\\T = m\left( {g + a} \right)\\\\ = \left( {{\rm{15}}{\rm{.0 Kg}}} \right)\left\{ {\left( {{\rm{9}}{\rm{.81 m}} \cdot {{\rm{s}}^{ - 2}}} \right) + \left( {{\rm{2}}{\rm{.91 m}} \cdot {{\rm{s}}^{ - 2}}} \right)} \right\}\\\\ = 190.8{\rm{ N}}\\\end{array}$

Ans: Part 1

The magnitude of the upward acceleration of the load of the bricks is $\left( {2.91{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)$ .

Part 2

The tension T in the rope, while the load is moving is $\left( {190.8{\rm{ N}}} \right)$ .