1.

30mn

Answer»

In the given figure .. applying force balance along x direct

T1cos45° = T2cos30°=> T1 = T2*(√3/2)/(1/√2) = √(3/2)T2.

and along y direction , it is.

mg = T1sin(45°) +T2sin(30°)=> mg = T1/√2 + T2/2

now putting the value of T1 in the 2nd equation.

=> √3T2/√2 +T2/2 = mg => T2(√6+1) = 2mg => T2 = 2mg/(√6+1)

and hence T1 = (√3/√2)*(2mg/(√6+1)) => T1 = √6mg/(√6+1).


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