Answer:
The pulling force F is 200 N
Explanation:
We note that the tension in a rope between two or three blocks is given by the relation;
Where we have;
m₁ → T₁ → m₂ → T₂ → m₃ → F
[tex]T_1 = \dfrac{m_1}{m_1 + m_2 + m_3} \times F[/tex]
[tex]T_2 = \dfrac{m_1 + m_2}{m_1 + m_2 + m_3} \times F[/tex]
m₁ = 2 kg
m₂ = 4 kg
m₃ = 4 kg
T₂ = 120 N
Therefore, by substitution, gives;
2 kg → T₁ → 4 kg → 120 N → 4 kg → F
[tex]120 \ N = \dfrac{2 \ kg + 4 \ kg}{2 \ kg+ 4 \ kg+ 4 \ kg} \times F = \dfrac{6 \ kg}{10 \ kg} \times F[/tex]
F = 120 N × (10 kg)/(6 kg) = 200 N
The pulling force F = 200 N.
