A block A weighing 100 kg rests on a block B and is tied with a horizontal string to the wall at C. Block B weighs 200kg. The coefficient of friction between A and B is 0.25 and between B and the surface is (1/3). The horizontal force P necessary to move the block B should be (g=10m/s2)

Question

A block A weighing 100 kg rests on a block B and is tied with a horizontal string to the wall at C. Block B weighs 200kg. The coefficient of friction between A and B is 0.25 and between B and the surface is (1/3). The horizontal force P necessary to move the block B should be (g=10m/s2)  (A) 1150 N (B) 1250 N (C) 1300 N (D) 1420 N

Tardigrade Admin · Accepted Answer

Friction between block A and block B and between block B and surface will oppose the P. ∴ P=FAB+FBS =μ ABmAg+μ BS(mA+mB)g =0.25× 100× 10+(1/3)(100+200)× 100 =1250 N

Q. A block A weighing 100 kg rests on a block B and is tied with a horizontal string to the wall at C. Block B weighs 200kg. The coefficient of friction between A and B is 0.25 and between B and the surface is 31​. The horizontal force P necessary to move the block B should be (g=10m/s2)

1150 N

1250 N

1300 N

1420 N

Friction between block A and block B and between block B and surface will oppose the P. ∴ P=FAB​+FBS​ =μAB​mA​g+μBS​(mA​+mB​)g =0.25×100×10+31​(100+200)×100 =1250N

Friction between block A and block B and between block B and surface will oppose the P.
$∴$ $P = F_{A B} + F_{BS}$
$= μ_{A B} m_{A} g + μ_{BS} (m_{A} + m_{B}) g$
$= 0.25 \times 100 \times 10 + \frac{1}{3} (100 + 200) \times 100$
$= 1250 N$