Consider the arrangement shown in fig.Friction coefficient for all the surfaces are shown in fig.and the rod connecting two upper blocks is horizontal.What is the minimum value F so that the rear two blocks start sliding with each other
Asked by m.nilu
14th August 2018,
10:26 PM
Answered by Expert
Answer:
acceleration a of system that combining all the four blocks is given by, a = F / (8m) ...............(1)
from the free body diagram given for small blocks, we have, μ×N = 2m×a .................(2)
N = 2m×g ................(3)
by substituting for N from eqn.(3) in eqn(2) and also by substituting for acceleration a using eqn.(1) in eqn.(2), we get F = μ×8m×g
since it is given μ=0.5, we need minimum force F = 4m×g
Answered by Expert
15th August 2018,
3:24 PM
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