2 Minor improvements. edit approved Jan 15 '14 at 12:24 user29727 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 : $$F_g cos(\theta) + m a_c sin(\theta) = F_n$$$$F_g \cos(\theta) + m a_c \sin(\theta) = F_n$$ $$F_g sin(\theta) = m a_c cos(\theta)$$$$F_g \sin(\theta) = m a_c \cos(\theta)$$ But if you see from ground frame the normal partly provides the centripetal force and also cancel outs the gravitational force i.e. $$F_n cos(\theta) = F_g$$$$F_n \cos(\theta) = F_g$$ $$F_n sin(\theta) = m a_c$$$$F_n \sin(\theta) = m a_c$$ In both frames as normal is involved in dealing with both forces it is greater than just the component of gravitational force. 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. 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. 1 answered Jan 12 '14 at 22:16 Rijul Gupta 3,91144 gold badges2323 silver badges5252 bronze badges 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.