A board of mass m = 1 kg lies on a table and a weight of M = 2kg on the board. What minimum force F (in N) must be applied on the board to pull it out from under the load? The coefficient of friction between the load and the board is μ1=0.25 and that between board and table is μ2=0.5 (Take g=10m/s2).
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answer is 22.5.
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Detailed Solution
f1max=μ1N1=0.25×20=5Nf2max=μ2N2f2=0.5(10+20)f2=0.5×30=15NWhen both board and block start sliding at F > 15 N, their acceleration will bea=F−153To start sliding between board and block we use5=2F−153⇒ F=22.5N