Question

The figure shows two 1.0kgblocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250g. The entire assembly is accelerated upward at 3.0m/s2 by force F?

What is F?

What is the tension at the top end of rope 1?

What is the tension at the bottom end of rope 1?
What is the tension at the top end of rope 2?

Answer 1

Concepts and reason

The concept of Newton’s second law of motion and tension is required to solve the problem.

The magnitude of the upward force acting on the two blocks is determined by finding the net force acting on the system, weight of the two blocks, and weight of each rope.

The tension at the top end of the rope 1 is determined by using the upward force and force acting on the block A due to gravity and due to upward force.

The tension at the bottom end of rope 2 is determined by using the tension at the top end of rope 1 and force acting on the rope due to gravity and due to upward force.

Tension at the top end of rope 2 is determined by using the tension at the at the top end of rope 1 and net force acting on the block B due to gravity and due to upward force.

Fundamentals

Newton’s second law of motion states that the net force acting on an object is directly proportional to the product of its mass and acceleration.

According to Newton’s second law, net force acting on an object is,

$F = ma$

Here, m is the mass of the object and a is the acceleration of the object.

Tension is a pulling force transmitted through a string, cord or wire. Tension is directed along the length of the wire and it is due to the forces pulling on the wire from either end.

Weight of an object is the force on an object due to gravity. Force acting on an object due to its weight is,

${F_{\rm{W}}} = mg$

Here, m is the mass of the object and g is acceleration due to gravity.

According to Newton’s second law of motion the net force acting on the system is,

$\begin{array}{c}\\{F_{{\rm{net}}}} = \left( {{m_1} + {m_2} + {m_3} + {m_3}} \right)a\\\\ = \left( {{m_1} + {m_2} + 2{m_3}} \right)a\\\end{array}$

Here, ${m_1}$ is the mass of block A, ${m_2}$ is the mass of block B, ${m_3}$ is the mass of rope, and a is the acceleration of the system.

The net force acting on the system is in upward direction and it is equal to,

${F_{{\rm{net}}}} = F - \left( {{m_1} + {m_2} + 2{m_3}} \right)g$

Here, F is the force acting in the upward direction and g is acceleration due to gravity.

Substitute $F - \left( {{m_1} + {m_2} + 2{m_3}} \right)g$ for ${F_{{\rm{net}}}}$ in equation ${F_{{\rm{net}}}} = \left( {{m_1} + {m_2} + 2{m_3}} \right)a$ and solve for F.

$\begin{array}{c}\\F = \left( {{m_1} + {m_2} + 2{m_3}} \right)g + \left( {{m_1} + {m_2} + 2{m_3}} \right)a\\\\ = \left( {{m_1} + {m_2} + 2{m_3}} \right)\left( {a + g} \right)\\\end{array}$

Substitute 1 kg for ${m_1}$ and ${m_2}$, 250 g for ${m_3}$, $3.00{\rm{ m/}}{{\rm{s}}^2}$ for a, and $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g in equation $F = \left( {{m_1} + {m_2} + 2{m_3}} \right)\left( {a + g} \right)$.

$\begin{array}{c}\\F = \left( {1{\rm{ kg}} + 1{\rm{ kg}} + 2\left( {250{\rm{ g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)} \right)\left( {3.00{\rm{ m/}}{{\rm{s}}^2} + 9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 32.0{\rm{ N}}\\\end{array}$

Consider a system consisting of block A.

According to Newton’s second law of motion the net force acting on block A is,

${F_{{\rm{net}}}} = {m_1}a$

Here, ${m_1}$ is the mass of block A, and a is the acceleration of the block.

The net force acting on the block is in upward direction and it is equal to,

${F_{{\rm{net}}}} = F - {T_1} - {m_1}g$

Here, F is the force acting in the upward direction, ${T_1}$ is the tension at the top end of the rope 1, and g is acceleration due to gravity.

Substitute ${m_1}a$ for ${F_{{\rm{net}}}}$ in equation ${F_{{\rm{net}}}} = F - {T_1} - {m_1}g$ and solve for ${T_1}$ .

$\begin{array}{c}\\{m_1}a = F - {T_1} - {m_1}g\\\\{T_1} = F - \left( {a + g} \right){m_1}\\\end{array}$

Substitute 32 N for F, 1 kg for ${m_1}$, $3.00{\rm{ m/}}{{\rm{s}}^2}$ for a, and $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g in equation ${T_1} = F - \left( {a + g} \right){m_1}$.

$\begin{array}{c}\\{T_1} = 32{\rm{ N}} - \left( {3.00{\rm{ m/}}{{\rm{s}}^2} + 9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {1{\rm{ kg}}} \right)\\\\ = 19.2{\rm{ N}}\\\end{array}$

Consider a system consisting of rope of mass ${m_3}$.

According to Newton’s second law of motion the net force acting on the rope is,

${F_{{\rm{net}}}} = {m_3}a$

Here, ${m_1}$ is the mass of block A, and a is the acceleration of the block.

The net force acting on the rope is in upward direction and it is equal to,

${F_{{\rm{net}}}} = {T_1} - {T_2} - {m_3}g$

Here, ${T_1}$ is the tension at the top end of the rope 1, ${T_2}$ is the tension at the bottom end of the rope 1, and g is acceleration due to gravity.

Substitute ${m_3}a$ for ${F_{{\rm{net}}}}$ in equation ${F_{{\rm{net}}}} = {T_1} - {T_2} - {m_3}g$ and solve for ${T_2}$ .

$\begin{array}{c}\\{m_3}a = {T_1} - {T_2} - {m_3}g\\\\{T_2} = {T_1} - \left( {a + g} \right){m_3}\\\end{array}$

Substitute 19.2 N for ${T_1}$, 250 g for ${m_3}$, $3.00{\rm{ m/}}{{\rm{s}}^2}$ for a, and $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g in equation ${T_2} = {T_1} - \left( {a + g} \right){m_3}$.

$\begin{array}{c}\\{T_2} = 19.2{\rm{ N}} - \left( {3.00{\rm{ m/}}{{\rm{s}}^2} + 9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {250{\rm{ g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right)\\\\ = 16.0{\rm{ N}}\\\end{array}$

Consider the system consisting of block B.

According to Newton’s second law of motion the net force acting on block B is,

${F_{{\rm{net}}}} = {m_2}a$

Here, ${m_2}$ is the mass of block B, and a is the acceleration of the block.

The net force acting on the block is in upward direction and it is equal to,

${F_{{\rm{net}}}} = {T_2} - {T_3} - {m_2}g$

Here, ${T_3}$ is the tension at the top end of the rope 2, ${T_2}$ is the tension at the bottom end of the rope 1, and g is acceleration due to gravity.

Substitute ${m_2}a$ for ${F_{{\rm{net}}}}$ in equation ${F_{{\rm{net}}}} = {T_2} - {T_3} - {m_2}g$ and solve for ${T_3}$ .

$\begin{array}{c}\\{m_2}a = {T_2} - {T_3} - {m_2}g\\\\{T_3} = {T_2} - \left( {a + g} \right){m_2}\\\end{array}$

Substitute 16.0 N for ${T_2}$, 1 kg for ${m_2}$, $3.00{\rm{ m/}}{{\rm{s}}^2}$ for a, and $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g in equation ${T_3} = {T_2} - \left( {a + g} \right){m_2}$.

$\begin{array}{c}\\{T_3} = 16.0{\rm{ N}} - \left( {3.00{\rm{ m/}}{{\rm{s}}^2} + 9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {1{\rm{ kg}}} \right)\\\\ = 3.20{\rm{ N}}\\\end{array}$

Ans:

The magnitude of the upward force is $32.0{\rm{ N}}$.

Tension at the top end of the rope 1 is $19.2{\rm{ N}}$.

Tension at the bottom end of the rope 1 is $16.0{\rm{ N}}$.

Tension at the top end of rope 2 is $3.20{\rm{ N}}$.