Two smooth cylindrical bars weighing W each lie next to each other in contact .A similar third bar is placed over the two bars .Neglecting friction the minimum horizontal force on each lower bar necessary to keep them together is
Asked by m.nilu
10th July 2018,
11:45 AM
Answered by Expert
Answer:
Figure shows how the normal forces are acting on bottom bars.
Triangle formed by connecting the 3 centers of circle is equilateral triangle.
From forces acting on vertical direction, we have 2×N×cos30 = w or 
F = N sin30 
Answered by Expert
16th July 2018,
3:49 PM
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