Question

Two blocks of masses
m = 5kg and m, = 6kg are
connected by a light string
passing over a light
frictionless pulley as shown.
The mass m, is at rest on the
inclined plane and mass m2
hangs vertically. If the angle
of incline (0) = 30', what is
magnitude and direction of
force of friction on the 5kg
block (g = 10ms 2)​

Answer 1

Answer

Explanation:

Force along plane, F

1

​

=mgsinθ=25N

Balancing forces along the plane;

∴T−mgsin30

0

−f=0

f=T−mgsin30

0

=60−25

=35 N

Answer 2

Answer:

For equilibrium condition

m2g= m1gsin30+f

60= 50/2 + f

f= 35N

friction force will act downward direction.

Hope it will help you