Question

Three blocks are stacked on top of each other inside anelevator as shown in the figure.Answer the following questions with reference to the eightforces defined as follows.1. the force of the 3 kg block on the 2 kg block,F_{3 on 2},2. the force of the 2 kg block on the 3 kg block,F_{2 on 3},3. the force of the 3 kg block on the 1 kg block,F_{3 on 1},4. the force of the 1 kg block on the 3 kg block,F_{1 on 3},5. the force of the 2 kg block on the 1 kg block,F_{2 on 1},6. the force of the 1 kg block on the 2 kg block,F_{1 on 2},7. the force of the 1 kg block on the floor, F_{1 onfloor and}8. the force of the floor on the 1 kg block,F_{floor on 1,}Assume the elevator is at rest. Rank themagnitude of the forces. Rank from largest to smallest.At first, I thought the magnitudes of eachforce were equal and have no force, because according toNewton's 1st law: If the vector sum of the forcesapplied to theobjects is zero, then the object would be either at rest or atconstant velocity (no acceleration).But then I realized that blocks 1 and 3 do nottouch, therefore zero force exert on each other. So, F_{3on 1} and F_{1 on 3} have thesmallest magnitude ofthe forces. (Magnitude cannot be below than zero).I think that F_{Floor on 1} has the largest magnitudebecause it balances the combined weights of all three blocks. So myanswer was F_{1 on floor}and F_{floor on 1},F_{3 on 2} and F_{2 on 3}, F_{2 on 1} andF_{1 on 2}, F_{3 on 1} andF_{1 on 3} And that was wrong.

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

is the same as when it would be at rest.

1. F(1 on the floor) &F(floor on 1)
2. F(1 on 2) &F(2 on 1)
3. F(2 on 3) &F(3 on 2)
4. F(3 on 1) &F(1 on 3)