Answer:
[tex]T_{B}=26.16 N[/tex]
Explanation:
We need to use the Newton's second law.
Block A
[tex]\Sigma F=m_{A}a[/tex]
[tex]T_{A}-m_{A}gsin(30)-\mu m_{A}gcos(30)=m_{A}a[/tex] (1)
Block B
[tex]\Sigma F=m_{B}a[/tex]
[tex]m_{B}g-T_{B}=m_{B}a[/tex] (2)
Pulley
Here we need to use Newton's law related to the torque.
[tex]T_{B}R-T_{A}R=I\alpha[/tex] (3)
I is the momentum of inertia of a cylinder (I=(1/2)mr²)
α is the angular acceleration (a/R)
Let's combine 1 and 2 to find T(B)-T(A)
[tex]T_{B}-T_{A}=m_{B}g-m_{A}g(sin(30)+\mu cos(30))-a(m_{A}+m_{B})[/tex] (4)
From 3 we have:
[tex]T_{B}R-T_{A}R=(1/2)m_{p}R^{2}*(a/R)[/tex]
[tex]T_{B}-T_{A}=(1/2)m_{p}*a[/tex] (5)
We can equal (4) and (5)
[tex](1/2)m_{p}*a=m_{B}g-m_{A}g(sin(30)+\mu cos(30))-a(m_{A}+m_{B})[/tex]
[tex](1/2)m_{p}*a+a(m_{A}+m_{B})=m_{B}g-m_{A}g(sin(30)+\mu cos(30))[/tex]
[tex]a((1/2)m_{p}+(m_{A}+m_{B}))=m_{B}g-m_{A}g(sin(30)+\mu cos(30))[/tex]
[tex]a=\frac{m_{B}g-m_{A}g(sin(30)+\mu cos(30))}{(1/2)m_{p}+(m_{A}+m_{B})}[/tex]
[tex]a=\frac{3*9.81-4*9.81*(sin(30)+0.15cos(30))}{(1/2)*4+(4+3)}[/tex]
[tex]a=0.52 m/s^{2}[/tex]
Therefore, using (2) we can find the tension in the rope between block B:
[tex]T_{B}=m_{B}g-m_{B}a[/tex]
[tex]T_{B}=m_{B}(g-a)[/tex]
[tex]T_{B}=27.87 N[/tex]
I hope it helps you!
