# How can vertical component of normal force balance the weight?

Consider a bicyle on a bank.

The weight of the bicycle acts downwards, we can resolve this vector to the normal force and a component down the bank. Hence the normal force is less than this weight force. However, now when we resolve the normal force vector to a vertical component and a horizontal component that produces centripetal acceleration, something confusing happens. The vertical component vector is smaller than the vector for weight force, but my textbook says that the vertical component of the normal force balances the weight force. All my diagrams show this cannot be the case. So how do you really show this?

• show your diagrams Commented Mar 9 at 14:25

The situation in your calculations describes the situation of the bicycle going straight in a banked road. The keyword is straight. With that you did not take friction into account. If you draw the diagrams without friction there will be a force pointing along the road which will make the bicycle go downwards. If you do not want that you will have to include a frictional force in the opposite direction with the same magnitude ($$mg sin \theta$$ if $$\theta$$ is the angle of the road with respect to the horizontal) this will supply the remaining part of the force ($$mg sin^2 \theta$$) needed to oppose the weight of the bicycle which along with the vertical force from the normal ($$mg cos^2 \theta$$) will oppose the weight. You will notice though that the force outside due to friction perfectly cancels out the force inwards due to the normal. Thus, the cycle will go in a straight path.

If you want to show the case where the cycle makes a turn you must understand that the normal force must be greater that the weight in magnitude as it has to both

1. keep the bycicle in the same height. Thus, opposing the entire force of gravity.
2. also provide the force needed to turn around the bank The vertical component of the normal equals the gravitational force and the horizontal component supplies the force for turning.

Hope this was helpful. I'm new so sorry if the formatting is off.

Consider the above figure, where $$m g$$ is the weight of the vehicle, $$N$$ is the normal reaction, $$f$$ is the frictional force, $$f \cos \left( \theta \right)$$ is the horizontal component of the frictional force, $$f \sin \left( \theta \right)$$ is the vertical component of the frictional force, $$N \cos \left( \theta \right)$$ is the vertical component of normal reaction and $$N \sin \left( \theta \right)$$ the horizontal component of the normal reaction.

The component $$N \cos \left( \theta \right)$$ of the normal reaction is balanced by the weight $$m g$$ of the vehicle and the component $$f \sin \left( \theta \right)$$ of the frictional force,

$$N \cos \left( \theta \right) = m g + f \sin \left( \theta \right) \therefore m g = N \cos \left( \theta \right) - f \sin \left( \theta \right) \tag{1}$$

Similarly, the horizontal component $$N \sin \left( \theta \right)$$ along with component $$f \cos \left( \theta \right)$$ of the frictional force provides the necessary centripetal force.

$$N \sin \left( \theta \right) + f \cos \left( \theta \right) = \frac{m v^{2}}{R} \tag{2}$$

Or

$$\frac{m v^{2}}{R} = N \sin \left( \theta \right) + f \cos \left( \theta \right) \tag{3}$$