Question

A 3 kg block (A) is placed on 6 kh block (B) which rests on the table . Coffecient of friction between A and B is 0.3 and between B and table is 0.6 . A 30N horizontal force is applied on block B. Calculate the friction force between A and B block

Answer 1

The friction force between A and B block will be zero.

Explanation:

Coefficient of friction of A block "μ" = 0.3

Coefficient of friction of B block "μ"  = 0.6

NA = 3 x g = 30 N

• F (limiting force) on A = 0.3 x 30 = 9 N

NB = (3 + 6) x 10 = 90 N

• F (limiting force) on B = 0.6 x 90 = 54 N

• Now this means the minimum force required to be applied on B for motion to occur is 54 N.

• But if we apply 30 N, motion won't occur.

• Friction force between A and B block will be zero and friction force between B and ground will be 30 N.

Hence the friction force between A and B block will be zero.

Also learn more

A 40kg slab rests on a frictionless floor . A 10 kg block rests on top of slab . coefficient of kinetic friction between the block and slab is 0.4 . A horizontal force of 100 N applied on 10 kg block .Find resulting acceleration of slab ?

https://brainly.in/question/1073898

Answer 2

Answer:

0

Explanation:

ANSWER is 0

bcz ...

there

is answer

is 0

.