A block $A$ of mass $100 k g$ rests on another block $B$ of mass $200 k g$ and is tied to a wall as shown in the figure. The coefficient of friction between $A$ and $B$ is $0.2$ and that between $B$ and the ground is $0.3$ . The minimum force $F$ required to move the block $B$ is ( $g = 10 m / s^{2}$ ):

900 N

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200 N

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1100 N

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700 N

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Solution

The correct option is C $1100 N$
Given : $μ_{1} = 0.2$ $f_{2} = 0.3$ $m_{B} = 200$ kg $m_{A} = 100$ kg
Frictional force acting between blocks A and B $f_{1} = μ_{1} m_{A} g$
$∴$ $f_{1} = 0.2 (100) (10) = 200$ $N$
Frictional force acting between block B and ground $f_{2} = μ_{2} (m_{A} + m_{B}) g$
$∴$ $f_{2} = 0.3 (100 + 200) (10) = 900$ $N$
Minimum force required to move block B, $F = f_{1} + f_{2}$
$⟹$ $F = 200 + 900 = 1100$ $N$

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Q. 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

(μ_{s})_{A} = 0.2

and the coefficient of static friction between block B and floor is

(μ_{s})_{B} = 0.3

. Then calculate the minimum value of force

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(g = 10 {m/s}^{2})

A block A of mass 200 kg rests on a block B of mass 300 kg. A is tied with a horizontal string to a wall. Coefficient of friction between A and B is 0.25 and that between B and floor is 0.2. The horizontal force F needed to move the block B is $(g = 10 m s^{- 2})$ .