Question

Find M (in kg) for which m2 will start sliding
over m1. Table on which m1 and m2 are
placed is smooth, u is the friction coefficient
between my and m2.
(Given : m1 = 2 kg, m2 = 1 kg, u = 0.5, g = 10
m/s2)
m2
m,
M​

Answer 1

Answer:

Gravitational Force on m

2

along the inclined plane in downward direction is,

F

G2

=m

2

gsin60

0

=1×g×

2

3

=

2

3

g

Similarly, Gravitational Force on m

1

along the inclined plane in downward direction is,

F

G1

=m

1

gsin30

0

=

3

×g×

2

1

=

2

3

g

So net external force on the tow-block system is F

G2

−F

G1

=0

So acceleration of the blocks is also zero.

So Tension in the string balances the gravitational force on blocks along the inclined plane.

So tension, T=F

G2

=F

G1

=

2

3

g

Explanation:

hope it is helpful for you

Answer 2

Answer:

gravitational force on M2