Answer:
a) [tex] F=32.0 N [/tex]
b) [tex] T_{1}=19.2 N [/tex]
c) [tex] T_{2}= 16.0 N [/tex]
d) [tex] T_{3}=3.2 N [/tex]
Explanation:
Let's do the free body diagram per each body. In this particular case the rope has a mass so we have:
First block
[tex] F-T_{1}-m_{1}g=m_{1}a [/tex] (1)
Frist rope
[tex] T_{1}-T_{2}-m_{r}g=m_{r}a [/tex] (2)
Second block
[tex] T_{2}-T_{3}-m_{2}g=m_{2}a [/tex] (3)
Second rope
[tex] T_{3}-m_{r}g=m_{r}a [/tex] (4)
When,
T
₁ is the tension at the top of the rope 1
T
₂ is the tension at the bottom of the rope 1
T
₃ is the tension at the top of the rope 2
Now, if we add all equations from (1) to (4), we will get the value of F,
[tex] F-(m_{1}+m_{2}+2m_{r})g=(m_{1}+m_{2}+2m_{r})a[/tex]
a) Solving this equation for F:
[tex] F=(m_{1}+m_{2}+2m_{r})(a+g)[/tex]
[tex] F=(m_{1}+m_{2}+2m_{r})(a+g)[/tex]
[tex] F=32.0 N[/tex]
b) Using the equation (1), we can find T₁
[tex] T_{1}=F-m_{1}(g+a)[/tex]
[tex] T_{1}=19.2 N[/tex]
c) Let's use the equation (2) to find T₂
[tex] T_{2}=T_{1}-m_{r}(g+a) [/tex]
[tex] T_{2}= 16.0 N [/tex]
d) Using the equation (4) we can find T₃:
[tex] T_{3}=m_{r}(a+g) [/tex]
[tex] T_{3}=3.2 N [/tex]
Have a nice day!
