Explanation:
Using Newtons second law on each block
F = m*a
Block 1
[tex]T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1[/tex]
Block 2
[tex]T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2[/tex]
Block 3
[tex]- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3[/tex]
Solving Eq1,2,3 simultaneously
Divide 1 and 2
[tex]\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4[/tex]
Put Eq 4 into Eq3
[tex]T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5[/tex]
Put Eq 5 into Eq2 and solve for a
[tex]a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6[/tex]
Substitute back in Eq2 and use Eq4 and solve for T2 & T1
[tex]T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\[/tex]
