As shown in Fig. 13–10a, the system is released from pulleys and the cord are neglected. Determine the force on a 20-kg block B.

Steps to Solve

1. Identify Forces Acting on Block B

The forces acting on the 20-kg block B are its weight ($W_B$) and the tension ($T$) in the cord. The weight can be calculated using the formula: $$ W_B = m_B \cdot g $$ where $m_B = 20 , \text{kg}$ and $g = 9.81 , \text{m/s}^2$.

2. Calculate Weight of Block B

First, calculate the weight of Block B: $$ W_B = 20 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 196.2 , \text{N} $$

3. Set Up the Equation of Motion

Assuming Block B is in free fall when released, the net force ($F_{net}$) acting on it can be given by: $$ F_{net} = W_B - T $$ By Newton's second law, this is also equal to: $$ F_{net} = m_B \cdot a $$ where $a$ is the acceleration of Block B.

4. Combine the Equations

Setting the two expressions for $F_{net}$ equal gives: $$ W_B - T = m_B \cdot a $$ This can be rearranged to find the tension: $$ T = W_B - m_B \cdot a $$

5. Determine the Acceleration

The problem does not specify another mass or pulleys' configuration. However, if Block B is in free fall with no other masses, its acceleration will be: $$ a = g = 9.81 , \text{m/s}^2 $$

6. Final Calculation of Tension

Substituting $a$ back into the tension formula gives: $$ T = 196.2 , \text{N} - 20 , \text{kg} \cdot 9.81 , \text{m/s}^2 $$ $$ T = 196.2 , \text{N} - 196.2 , \text{N} = 0 , \text{N} $$