Question

In the given figure, the coefficient of static friction
between the block B and horizontal table is
0.4. What should be the maximum mass of hanging
block A, so the system remain in equilibrium is
(g = 10 m/s2)​

Answer 1

Answer:

Explanation:

Hope it will help you

Answer 2

Answer:

THE ANSWER TO THIS QUESTION IS FOUR KILOGRAM

Explanation:

Using the concept of balancing forces on the entire system i.e using the concepts of NEWTONIAN MECHANICS

We start balancing the forces on each of the two blocks connected by a string which is in extensible which indicates ideal conditions and thus the tension is uniform all throughout the string

Considering the block hanging,

The weight of the block is balanced by the tension in the string

Hence,

T(tension)=M×g where M is the mass of hanging block  → 1

Considering the other block

Tension of the string is balanced by the frictional force

T=μmg → 2

where m=mass of block

           μ=coefficient of static friction

from 1 and 2 equating both the tensions we obtain

Mg=μmg

M=μm=0.4×10=4 kg