8.7 Free body diagram (application)  (Page 27/3)

Problem : Draw free body diagram of three blocks placed one over other as shown in the figure.

Solution : We start drawing FBD of the topmost block as we do no need to consider normal force due to an overlying body as the case with other blocks.

The forces on the block “C” are :

• $W_C=m_{C}g=$ its weight, acting downward
• $N_B=$ normal reaction on “C” due to the upper surface of block B, acting upward

The forces on the block “B” are :

• $W_B=m_{B}g=$ its weight, acting downward
• $N_B=$ normal reaction on “B” due to the lower surface of block C, acting downward
• $N_A=$ normal reaction on “B” due to the upper surface of block A, acting upward

The forces on the block “A” are :

• $W_A=m_{A}g=$ its weight, acting downward
• $N_A=$ normal reaction on “A” due to the lower surface of block B, acting downward
• $N_O=$ normal reaction on “A” due to horizontal surface, acting upward

The FBD of the blocks as points with external forces are shown here. Note that motion is not involved. Hence, no information about acceleration is given in the drawing.

We have deliberately not shown the coordinate system which may be selected, keeping in mind the inputs available. In this instant case, however, a vertical axis is clearly the only choice.