The correct option is
A direction of force of friction is up the plane.
By balancing force on
10 kg mass in perpendicular direction to the incline we get
N=mgcos37∘
⇒N=100×45⇒N=80 N
Now we know friction is given by
f=μN=0.7×80⇒f=56 N
By balancing force on
4 kg mass we get
T−40=4a .....(i)
where a is the acceleration with which it is going upward
Now balancing force on
10 kg block in direction along the incline we get
100sin37∘−T−f=10a ....(ii)
Acceleration of both the blocks will be same.
From (i) and (ii)
f=20−14a .....(iii)
From (iii) we can conclude that
as we increase the magnitude of
a, the value of friction is decreasing.
So
f≤20
Hence maximum value of friction can be
20
And
T can be calculated from force balancing of
4kg block
T=40N
So only option (a) is correct as friction acts in the direction opposite to the tendency of slipping