Chapter 4, Problem 30P

Figure 4-53 [shows a block (mass m_A) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (m_B), which hangs vertically. (a) Draw a free-body diagram for each block, showing the force of gravity on each, the force (tension) exerted by the cord, and any normal force.(b) Apply Newton's second law to find formulas for the acceleration of the system and for the tension in the cord. Ignore friction and the masses of the pulley and cord.

Figure 4-53
Problems 32 and 33. Mass m_Arests on a smooth horizontal surface; m_Bhangs vertically.

To determine

(a) To Determine:

Free body diagram of block A and block B

Answer to Problem 30P

Solution:

Explanation of Solution

The block A is kept on a horizontal surface. Hence horizontal surface applies normal force in an upward direction. The cord connected on right applies tension force in the right direction. The force of gravity act in down direction.

The block B is hanging from the cord. Hence cord applies a tension force on the block in upward direction and force of gravity by earth acts in down direction.

Tension force $=T$

Weight $=mg$

Acceleration $=a$

Normal force $=F_n$

To determine

b)

The acceleration of the system and tension in the cord.

Answer to Problem 30P

Solution:

$a=\frac{m_{B}g}{(m_A+m_B)}$

$T=\frac{m_A{m}_{B}g}{(m_A+m_B)}$

Explanation of Solution

Due to tension force in right direction, the block A on the horizontal surface accelerates in the right direction.

The net force on the block B acts in down direction and hence block B accelerate downward

Given:

Tension force $=T$

Normal force $=F_n$

Acceleration $=a$

Mass of block A $=m_A$

Mass of block B $=m_B$

Formula Used:

$F=ma$

Calculation:

For block A:

Force equation along the horizontal direction is given as

$T=m_{A}a$            eq-1

For block B:

Force equation along the vertical direction is given as

$m_{B}g-T=m_{B}a$

Using eq-1

$\begin{array}{l}{m}_{B}g-m_{A}a=m_{B}a\\ a=\frac{m_{B}g}{(m_A+m_B)}\end{array}$            eq-2

Using eq-1

$T=m_{A}a$

Using eq-2

$T=\frac{m_A{m}_{B}g}{(m_A+m_B)}$