    Question

# Mass of block A is 100 kg and that of block B is 200 kg. As shown in figure, the block A is attached to a string tied to the wall. The coefficient of static friction between blocks A and B is (μs)A=0.2 and the coefficient of static friction between block B and floor is (μs)B=0.3. Then calculate the minimum value of force F required to move the block B. (g=10 m/s2) A
980 N
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B
1000 N
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C
1100 N
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D
1200 N
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Solution

## The correct option is C 1100 N From the FBD of A: T=f1........(1) From FBD of B: F=f1+ f2......(2) Since we want the minimum value of F (for just slipping condition), we will take the limiting values of friction. f1=(μs)A mA g .........(3) and f2=(μs)B(mA+mB)g ......(4) Substituting (3) and (4) in (2), F=f1+f2 =(0.2)(100g)+(0.3)(300g)=110 g=1100 N  Suggest Corrections  0      Related Videos   Rubbing It In: The Basics of Friction
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