Two blocks with masses m1 and m2 of 10kg and 20kg respec are placed as in fig. mew=0.2 bet all surfaces then tension in string and acceleration of m2 block at this moment will be
Asked by m.nilu
15th August 2018,
5:23 PM
Answered by Expert
Answer:
Free body diagram of both the blocks are given in figure.
From the m1 block, forces acting on vertical direction, we have, T sin30 = 10g ............(1)
from eqn.(1), we get Tension in the string, T = 196 N
Horizontal componenet of Tension of the string pressing m2 block.
Hence normal reaction force N = Tcos30 = 196×(√3/2) = 170 N
for the m2 block, using the vertical direction forces, Newton's law is written as :
20×g - 2×μ×N = 20×a ....................(2)
by substituting values for μ=0.2, N=170 , we get a = 6.4 m/s2
Answered by Expert
18th August 2018,
6:37 PM
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