Question-53727

Question Number 53727 by Tawa1 last updated on 25/Jan/19

Answered by Kunal12588 last updated on 25/Jan/19

j u s t t r y i n g

T_{1}, T_{2}, T_{3}, T_{4} - l o w e r c a b l e s

T_{1} = T_{2} = T_{3} = T_{4} = k (i d e n t i c a l c a b l e s)

\frac{1}{2} (\sqrt{2^{2} + 0 . 5^{2}}) = h

{t a n}^{- 1} (2 / h) = ϕ

4 k c o s ϕ - m g = 0

4 k c o s ϕ = w

k = \frac{4.9}{4 c o s ϕ}

T_{u p p e r} = 4 k c o s ϕ [n o t s o l v e d i t b c u z i t h i n k i t s w r o n g]

p l e a s e r e p o r t w i t h a n s w e r

Commented by Tawa1 last updated on 25/Jan/19

God bless you sir

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

d i m e n s i o n o f b o x = l \times b \times h

d i a g o n a l = \sqrt{l^{2} + b^{2}} =

t a n θ = \frac{h}{\frac{\sqrt{l^{2} + b^{2}}}{2}} = \frac{2 h}{\sqrt{l^{2} + b^{2}}}

4 T_{l o w e r c a b l e} s i n θ = m g

T_{l o w e r c a b l e} = \frac{m g}{4 s i n θ}

T_{u p p e r} = m g

Commented by Tawa1 last updated on 25/Jan/19

God bless you sir . you can help me complete it if correct sir .

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