1.

A solid sphere of mass 10 kg is placed over two smooth inclined placed as shown in figure. The normal reactions at 2 is (g = 10 m//s^(2))

Answer»

50
30
20
70

Solution :
`N_(1) sin 30^(@) = N_(2) sin 60^(@)`
`N_(1) cos 30^(@) + N_(2) cos 60^(@) = MG`
Solving above equation
`N_(2) = (mg)/(2) = (10 xx 10)/(2) = 50`


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