Question

A block of mass m1=3.7 kg on a frictionless plane inclined as angle θ=30 degrees is connected by a cord over a massless, frictionless pulley to a second block of mass m2=2.3 kg hanging vertically (shown above). What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?

Answer 1

Concepts and reason

The concepts used to solve this problem are newton’s second law, force of tension, and resolution of forces into their components.

Resolve the gravitational force acting on the block into its components.

Apply newton’s second law to each block.

Use the equations of motion of the two blocks to find their acceleration.

Finally find the tension in the cord using the value of acceleration.

Fundamentals

Newton’s second law states that for a body, “the rate at which its momentum changes is proportional to the net force acting on the body”.

Expression for newton’s second law is,

$\vec F = m\vec a$

Here, $\vec F$ is the net force acting on a body, $m$ is the mass of the body, and $\vec a$ is the acceleration produced on it.

Any force can be resolved into its components, which acts along each axis of the coordinate system under consideration.

The force is resolved into vertical and horizontal components such as,

${m_1}g\cos \left( \theta \right)$ and ${m_1}g\sin \left( \theta \right)$

(a)

The forces, that are acting downwards, are considered to be positive and that are acting upwards are considered to be negative.

Block of mass ${m_1}$ is on the inclined plane.

This mass experiences a force of gravity which is expressed as,

${\vec F_g} = {m_1}\vec g$

Here, ${\vec F_g}$ is the gravitational force, ${m_1}$ is the mass of block on the inclined plane, and $\vec g$ is the acceleration due to gravity.

The net force acting on the block of mass ${m_1}$ is,

${m_1}a = T - {m_1}g\sin \theta$ …… (1)

Here, a is the acceleration of each block, T is the tension, and $\theta$ is the angle.

The net force acting on the block of mass ${m_2}$ is,

${m_2}a = {m_2}g - T$ …… (2)

Rewrite the equation (2) in terms of T,

$T = {m_2}g - {m_2}a$ …… (3)

Substitute equation (3) in (1).

${m_1}a = \left( {{m_2}g - {m_2}a} \right) - {m_1}g\sin \theta$

Rearranging the above expression in terms of a.

$\begin{array}{c}\\a\left( {{m_1} + {m_2}} \right) = g\left( {{m_2} - {m_1}\sin \theta } \right)\\\\a = \frac{{g\left( {{m_2} - {m_1}\sin \theta } \right)}}{{\left( {{m_1} + {m_2}} \right)}}\\\end{array}$

Substitute $9.8\,{\rm{m}}{{\rm{s}}^{ - 2}}$ for$g$, $3.7\,{\rm{kg}}$ for${m_1}$, $30^\circ$ for $\theta$ and $2.3\,{\rm{kg}}$ for ${m_2}$ in the above expression.

$\begin{array}{c}\\a = \frac{{\left( {9.8\,{\rm{m}}{{\rm{s}}^{ - 2}}} \right)\left( {2.3\,{\rm{kg}} - 3.7\,{\rm{kg}}\sin \left( {30^\circ } \right)} \right)}}{{\left( {3.7\,{\rm{kg}} + 2.3\,{\rm{kg}}} \right)}}\\\\ = 0.73\,{\rm{m}}{{\rm{s}}^{ - 2}}\\\end{array}$

Therefore, the acceleration experienced by each block is $a = 0.73\,{\rm{m}}{{\rm{s}}^{ - 2}}$.

(b)

The net acceleration of the block with mass ${m_2}$ is expressed as,

${m_2}a = {m_2}g - T$

The tension is found to be lesser in magnitude than the force of gravity on the hanging block.

Hence, the net acceleration is acting downwards.

(c)

The tension in the cord can be calculated using the expression (3).

$T = {m_2}g - {m_2}a$

Substitute $2.3\,{\rm{kg}}$ for ${m_2}$, $9.8\,{\rm{m}}{{\rm{s}}^{ - 2}}$, and $0.73\,{\rm{m}}{{\rm{s}}^{ - 2}}$ for $a$.

$\begin{array}{c}\\T = \left( {2.3\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m}}{{\rm{s}}^{ - 2}}} \right) - \left( {2.3\,{\rm{kg}}} \right)\left( {0.73\,{\rm{m}}{{\rm{s}}^{ - 2}}} \right)\\\\ = 21\,{\rm{N}}\\\end{array}$

Hence, the tension in the cord is $T = 21\,{\rm{N}}$.

Ans: Part a

The acceleration experienced by each block is $a = 0.73\,{\rm{m}}{{\rm{s}}^{ - 2}}$.

Part b

The net acceleration of the hanging block is downwards.

Part c

The tension in the cord is $T = 21\,{\rm{N}}$