Question

A 4 kg block A is placed at the top of 8 kg block B which rests on a smooth table. A just slips on B when a force 20 N is applied on A. The maximum horizontal force F required to make both A & B move together is​

Answer 1

Answer:

ANSWER

When they both move together

a=

TotalMass

F

=

12

F

For A

F−12=4×

12

F

⟶Newtons 2nd Law

∴F−

3

F

=12

F=18N

Answer 2

Explanation:

Given A 4 kg block A is placed at the top of 8 kg block B which rests on a smooth table. A just slips on B when a force 20 N is applied on A. The maximum horizontal force F required to make both A & B move together is

• So A is a 4 kg block is placed on block B at the top of 8 kg block.
• So let mass of block A = m and mass of block B = M
• So A will be the limiting frictional force.
• So limiting frictional force of block A, f = 20 N
• Let acceleration of block A and B = a
• So there are  directions from A that will be f , mg and N
• Now N = mg
• So force f = ma
• Or μ mg = ma where μ is the coefficient of friction between block A and B
• So there are directions from B that will be F, Mg and N’
•         So N’ = Mg + N
•                    = Mg + mg
•        F – f = M a
•       Or F = f + m a
•                                 ( f = m a = 20  
•                                    Or m a = 20
•                                         a = 20 / m)
•      F = 20 + M (20 / m)
•         = 20 + 8 x 20 / 4
•        = 20 + 40
•        = 60 N
• So the maximum force will be 60 N

Reference link will be

https://brainly.in/question/1073919