Explain, please?

Lets' assume that the distances are from end A.

Taking clockwise moments about A we have:

(4 x 12) + (6 x 24) + (2 x 30) + (8 x 36) + (10 x 48)

i.e. 48 + 144 + 60 + 288 + 480 => 1020

T₂ has an anticlockwise moment of T₂ x 60

so, 1020 = 60T₂

=> T₂ = 17 Newtons

Now, T₁ + T₂ = 30

Hence, T₁ = 13 Newtons

:)>     

It seems that (b) is the correct answer.
E is just the rod so it will be half way at 60/2 = 30cm
We know the sum of the forces down = sum of forces up (i.e. T1+T2)
So T1+T2=4+6+2+8+10=30T1+T2=30
Also we know that the moments must also be equal (i.e. force x distance to pivot)
If we use A as our origin or pivot point then (Cx4)+ (Dx6) etc =T2 x 60
So get  (4x12) + (6x24) + (2x30) + (8x36) +(10x48) = T2 x 60
Adding it up gets 1020=60T2
or T2=1020/60=17
So sub in first equation to get T1+17=30
So T1=30-17=13
Finally, T1=13 and T2=17

A statics problem - sum of forces = zero = sum of moments

ΣF = 0
4 + 6 + 8 + 10 + 2 = T₁ + T₂ = 30
All of the possible answers satisfy this condition, so the solution will be found in the sum of moments.

ΣM = 0
Assuming the rod has a uniform linear density and taking moments about point A, T₁ contributes no moment. The moments of the other loads are, assuming the rod has a uniform linear density:
60T₂ = 4*12 + 6*24 + 8*36 + 10*48 + 2*30
T₂ = 12(4 + 12 + 24 + 40 + 5)/5
T₂ = 85/5 = 17
T₁ = 30 - 17 = 13

Ans: (b) T₁ = 13, T₂ = 17

4 6 8 10

   C D F ad G are supported in inverse proportio from distance to the end
    C  12/60  supported by the Right end  
     D   24/60 supported by the Right end  
   etc   (12/60) X 4  +  (24/60) X 6  +  (36/60) X 8  +  (48/60) X 10  + 1  =  17 N  =  T2   ....  so answer is b