Given:
• Mass of block A = 6.0 kg
,
• Mass of block B = 7.0 kg
,
• Mass of block C = 13.0 kg
,
• Force, F = 13.0 N
Let's find the magnitude of the tension in the rope between B and C.
Let's first find the acceleration.
We have:
[tex]13-T_B+T_B-T_A+T_A=6a+7a+13a[/tex]
Thus, we have:
[tex]\begin{gathered} 13=26a \\ \\ a=\frac{13}{26} \\ \\ a=0.5\text{ m/s}^2 \end{gathered}[/tex]
To find the tension between blocks B and C, we have the equation:
[tex]\begin{gathered} F-T_B=M_C*a \\ \\ T_B=F-M_c*a \end{gathered}[/tex]
Where:
F = 13 N
Mc is the mass of block C = 13 kg
a is the acceleration = 0.5 m/s²
Thus, we have:
[tex]\begin{gathered} T_B=13-13*0.5 \\ \\ T_B=13-6.5 \\ \\ T_B=6.5\text{ N} \end{gathered}[/tex]
Therefore, the magnitude of the tension in the rope between blocks B and C is 6.5 N
ANSWER:
6.5 N
