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Question

# In the figure shown below, a horizontal force F is applied on 6 kg block towards left. If the coefficients of friction between the surfaces are 0.3 and 0.4 as shown in the figure. then the value of tension in the rope and force required just to slide the 6 kg block under the 3 kg block is (Take g=10 m/s2)

A
T=3 N,F=40 N
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B
T=6 N,F=45 N
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C
T=9 N,F=45 N
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D
T=9 N,F=40 N
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Solution

## The correct option is C T=9 N,F=45 NAs per the question, the block of mass 6 kg will just slide under block of mass 3 kg. Then, the condition of friction will be limiting. The FBDs of the blocks are as shown At equilibrium, we have N1=3g=30 N f1=T In limiting case, f1=μ1N1=0.3×30=9 N ⇒ T=9 N Again, from the FBD of 6 kg block, we have N2=N1+6g=30+60=90 N F=f1+f2 In limiting case, f2=μ2N2=0.4×90=36 N ⇒ F=9+36=45 N

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