Two equal masses of m kg each are fixed at the vertices
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A weightless rod of length 2 l carries two equal masses m, one secured at lower end A and the other at the middle
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The angle between GC and positive X-axis is 30 and so is the angle between GB and negative X-axis.
On resolving the forces on 2m due to masses at B and C along X-axis :
Mode (Fx net) = mode (F cos 30 i - F cos 30 i) = 0
Mode (Fy net) = mode (Fj -(Fsin30+Fsin30)j )
= mode (Fj -Fj) = 0
Hence,
Fnet = Resultant force on mass at G due to masses at A, B and C = √(Fx net)2+(Fy net)2
Fnet = 0
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