Question

Part B

Now, assume the elevator is moving upward at increasing speed. Rank the magnitude of the forces.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Answer the following questions with reference to the eight forces defined as follows.
the force of the 3kg block on the 2kg block, F3on2
the force of the 2kg block on the 3kg block, F2on3
the force of the 3kg block on the 1kg block, F3on1
the force of the 1kg block on the 3kg block, F1on3
the force of the 2kg block on the 1kg block, F2on1
the force of the 1kg block on the 2kg block, F1on2
force of the 1kg block on the floor, F1onfloor
force of the floor on the 1kg block, Fflooron1

Please note: Before this question they had me rank them the same way except if the elevator was at rest, I know the correct ranking of that scenario going from largest to smallest is 1kg = force by floor on 1kg > force of 2 kg block on 1 kg block = force of 1 kg block on 2 kg block >force of 2 kg block on 3 kg = force of 3 kg block on 2 kg > force of 3 kg block on 1 kg block = force of 1 kg block on 3 kg block

Answer 1

Concepts and reason

The concepts required to solve this problem are Newton’s second law and third law.

Calculate the forces on each block by using the Newton’s second law and third law. Arrange the calculated forces in descending order.

Fundamentals

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,

$\sum {F = ma}$

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

Newton’s third law is as follows:

${F_{12}} = {F_{21}}$

Here, ${F_{12}}$ is force applied by first body on second body and ${F_{21}}$ is force applied by second body on first body.

Calculate force of 3 kg block on 2 kg block.

The apparent weight of 3 kg block is ${F_{32}}$.

Use the Newton’s second law on block of 3 kg mass.

$\begin{array}{c}\\ma = {F_N} - mg\\\\{F_N} = m\left( {g + a} \right)\\\\ = \left( {3{\rm{kg}}} \right)\left( {g + a} \right)\\\end{array}$

Here, ${F_N}$ is same as apparent weight of the 3 kg block.

According to Newton’s third law ${F_{32}} = {F_{23}}$.

Since, block of 3 kg is not in contact with block of 1 kg, there is no force acting, that is,

${F_{31}} = {F_{13}} = 0{\rm{ N}}$

Calculate the force of the 2 kg block on the 1 kg block.

Use the Newton’s second law on 2 kg block.

$\begin{array}{c}\\\left( {{m_2} + {m_3}} \right)a = {F_N} - \left( {{m_2} + {m_3}} \right)g\\\\{F_N} = \left( {{m_2} + {m_3}} \right)\left( {g + a} \right)\\\\ = \left( {5{\rm{kg}}} \right)\left( {g + a} \right)\\\end{array}$

Here, ${F_N}$ is normal force acting on 2 kg block.

Thus, the force of the 2 kg block on the 1 kg block is,

${F_{21}} = 5\left( {g + a} \right)$

According to Newton’s third law,

${F_{21}} = {F_{12}}$

Calculate the force of 1 kg block on the floor as follows,

Use the Newton’s second law on 1kg block.

$\begin{array}{c}\\\left( {{m_1} + {m_2} + {m_3}} \right)a = {F_N} - \left( {{m_1} + {m_2} + {m_3}} \right)g\\\\{F_N} = \left( {{m_1} + {m_2} + {m_3}} \right)\left( {g + a} \right)\\\\ = 6\left( {g + a} \right)\\\end{array}$

The ${F_N}$ is the normal force of 1 kg block.

Thus, the force of 1 kg block on floor is,

${F_{10}} = 6\left( {g + a} \right)$

According to Newton’s third law,

${F_{10}} = {F_{01}}$

Order of magnitude of force from the forces calculated in step 1 is as follows:

${F_{10}} = {F_{01}} > {F_{21}} = {F_{12}} > {F_{32}} = {F_{23}} > {F_{31}} = {F_{13}}$

Ans:

Order of magnitude of force is ${F_{10}} = {F_{01}} > {F_{21}} = {F_{12}} > {F_{32}} = {F_{23}} > {F_{31}} = {F_{13}}$.

Answer 2

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