An object of mass $m$ on an incline with friction experiences the following three forces:
- Gravity
- Normal force
- Friction
The normal force points away from the incline's surface and is perpendicular to it. The force of friction is parallel to the incline and points in the direction opposing the motion of the object (in this case that means it points up the incline). The gravitational force points down (in the direction of the negative $y$-axis).
Since the object is not accelerating in the direction perpendicular to the incline (otherwise it would be falling through the surface or losing contact with the surface), one concludes that the component of the gravitational force perpendicular to the incline cancels the normal force, and the net force in that direction is zero.
What remains is the component of the gravitational force pointing down the incline, and the friction force pointing up the incline. A picture and some triangle-drawing/trig should convince you that the magnitude of the component of the gravitational force pointing down the incline is
$m g \sin (30^\circ) = (1\,\mathrm{kg})(9.8\,\mathrm{m}/\mathrm{s}^2)(1/2) = 4.9\,\mathrm N$.
The friction force pointing up the incline has magnitude $1.5\,\mathrm N$. Since the component of the gravitational force along the incline points down the incline, and the friction force points up the incline, the magnitude of the net force is just
$4.9\,\mathrm N - 1.5\,\mathrm N = \boxed{3.4\,\mathrm N}$.
Let me know if anything here was confusing, and I can make more comments.
Cheers!
You have asked what causes the increase in force from the earlier case, try seeing it from the body's frame as it will be in rest there. The body exerts a centrifugal force on the plane along with gravitational force, the resultant of these two forces is matched up by the normal provided by the banked road, now since the resultant of centrifugal force and gravitational force is more than gravitational force alone, the normal force increases in magnitude.
In terms of equation :
$F_g \cos(\theta) + m a_c \sin(\theta) = F_n$
$ F_g \sin(\theta) = m a_c \cos(\theta)$
But if you see from ground frame the normal partly provides the centripetal force and also cancel outs the gravitational force i.e.
$F_n \cos(\theta) = F_g$
$F_n \sin(\theta) = m a_c$
In both frames as normal is involved in dealing with both forces it is greater than just the component of gravitational force.
Best Answer
Because friction is always parallel to the bank surface and a normal force is always perpendicular to it.
Imagine the extreme scenarios:
With the bank being flat (no banking), $\theta=0$, a normal force would point straight up and wouldn't be aiding to the horizontal centripetal force that causes turning.
With the bank being vertical, $\theta=90^\circ$, the friction can only be vertical and thus can't aid in providing the horizontal centripetal force.
At any angle in between, both forces contribute a little to the horizontal centripetal force. The steeper the bank, the larger the normal force and smaller the friction.