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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.

Relevant knowledge

If a body moves at an acceleration

*a*relative to the acceleration due gravity*g*then its relative acceleration will be given as......... ADVERTISEMENT .........

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This problem can be solved using Newton’s second and third laws.

Use the Newton’s second and third laws to calculate the forces for each block. Then arrange the calculated forces in descending or ascending order.

Newton’s second Lawstates, that the net force exerted on an object by a mass and its final acceleration is equal. The equation of Newton’s second law,[katex]\sum {F = ma} [/katex]

Here, [katex]\sum F [/katex]

Newton’s third Lawis:[katex]{F_{12}} = {F_{21}}[/katex]

Here, [katex]{F_{12}}[/katex]

Calculate the force of a 3kg block on a 2kg block.

The apparent weight for a 3kg block is [katex]{F_{32}}[/katex].

Apply Newton’s second law to block mass of 3 kg.

[katex]\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}[/katex]

Here, [katex]{F_N}[/katex]

Newton’s third law [katex]{F_{32}} = {F_{23}}[/katex]

Block of 3kg is not in contact block of 1kg, so there is no force acting.

[katex]{F_{31}} = {F_{13}} = 0{\rm{ N}}[/katex]

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

Apply Newton’s second law to 2kg blocks

[katex]\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}[/katex]

Here, [katex]{F_N}[/katex]

The force of the 2kg block on the 1kg block is thus,

[katex]\sum F [/katex]1

Newton’s third law states that

[katex]\sum F [/katex]2

As follows: Calculate the force of 1kg block on the floor.

Apply Newton’s second law to 1kg blocks

[katex]\sum F [/katex]3

The [katex]{F_N}[/katex]

The force of 1kg block on the floor is therefore,

[katex]\sum F [/katex]5

Newton’s third law states that

[katex]\sum F [/katex]6

The following is the order of magnitude of force calculated from step 1.

[katex]\sum F [/katex]7Ans:

[katex]\sum F [/katex]7 is the order of magnitude of force