Chapter 4, Problem 59A

To determine

To Compare: The normal force acting between block and the ground in different cases and arrange them in descending orders.

Answer to Problem 59A

  $N_3>N_1>N_2$

Explanation of Solution

Formula used:

From Newton’s second law of motion, if the blocks are in equilibrium, then:

  $F_{net}$ = 0 N

Calculation:

Let, mass of block is $M$ and mass of another block is $m$ .

The free body diagram of the blocks is shown as,

  

The block is at rest so net force acting on the block will be zero.

The force equations are,

  $N_1=Mg\text{ (1)}$

In second situation, the normal force on block is provided by spring and that spring force will be equal to normal force ( $N_2$ ). So, the force equation for second situation is,

  $\begin{array}{l}{N}_2+T=Mg\\ N_2=Mg-T\text{ (2)}\end{array}$

  $N_3=(M+m)g\text{ (3)}$

On comparing the normal force from equations $(1)$ , $(2)$ and $(3)$ the order of normal force between the block and ground is $N_3>N_1>N_2$

Conclusion:

The order of normal force between the block and ground is $N_3>N_1>N_2$ .