Explain, please?

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  • 6 months ago
    Favorite Answer

    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

    :)>     

  • 6 months ago

    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

  • TomV
    Lv 7
    6 months ago

    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

  • david
    Lv 7
    6 months ago

    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

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