Question

the blocks in figure 8 are pulled to the right and accelerate with a magnitude of 2.1 m/s². if ( m_1 = 4.0 ) kg, ( m_2 = 2.3 ) kg, and ( m_3 = 3.4 ) kg, calculate the tension in each string. assume that friction is negligible.

Explanation:

Step1: Analyze Tension in String C

To find the tension in string C, we consider only \( m_3 \). Using Newton's second law \( F = ma \), the tension \( T_C \) provides the force to accelerate \( m_3 \).
\( T_C = m_3 \times a \)
Substitute \( m_3 = 3.4 \, \text{kg} \) and \( a = 2.1 \, \text{m/s}^2 \):
\( T_C = 3.4 \times 2.1 = 7.14 \, \text{N} \)

Step2: Analyze Tension in String B

For string B, we consider \( m_2 \) and \( m_3 \) as a system. The tension \( T_B \) accelerates both \( m_2 \) and \( m_3 \).
\( T_B = (m_2 + m_3) \times a \)
Substitute \( m_2 = 2.3 \, \text{kg} \), \( m_3 = 3.4 \, \text{kg} \), and \( a = 2.1 \, \text{m/s}^2 \):
\( m_2 + m_3 = 2.3 + 3.4 = 5.7 \, \text{kg} \)
\( T_B = 5.7 \times 2.1 = 11.97 \, \text{N} \)

Step3: Analyze Tension in String A

For string A, we consider \( m_1 \), \( m_2 \), and \( m_3 \) as a system. The tension \( T_A \) accelerates all three masses.
\( T_A = (m_1 + m_2 + m_3) \times a \)
Substitute \( m_1 = 4.0 \, \text{kg} \), \( m_2 = 2.3 \, \text{kg} \), \( m_3 = 3.4 \, \text{kg} \), and \( a = 2.1 \, \text{m/s}^2 \):
\( m_1 + m_2 + m_3 = 4.0 + 2.3 + 3.4 = 9.7 \, \text{kg} \)
\( T_A = 9.7 \times 2.1 = 20.37 \, \text{N} \)

Answer:

Tension in string A: \( \boldsymbol{20.4 \, \text{N}} \) (rounded), Tension in string B: \( \boldsymbol{12.0 \, \text{N}} \) (rounded), Tension in string C: \( \boldsymbol{7.14 \, \text{N}} \)