Normal forces are a special type of force that are only present for an object on some surface that "neutralize" the force of gravity and related forces (though often it is just gravity). So, if an object is suspended in air, there is no normal force. The normal force is very important in the physical world, as the first example demonstrates:
Example 1: A block is resting on a table. If there was no normal force, the only force acting on the block would be gravity, meaning it would rip through the table and everything beneath it until it reached the center of the earth.
If that isn’t scary enough, remember that this block could easily be a person! However, now that we realize why the normal force is important, the question remains: how do we calculate the normal force?
The magnitude of the normal force depends on the other forces acting on the object. The calculations on the normal force depend on these other forces. Gravity will always be one of these forces. As a reminder, for any object on Earth with a mass of \(m\),
$$F_g = mg$$
\(g\) represents the gravity acceleration constant on earth. Assume all examples in this article occur on earth, so \(g = 9.8 \; \frac{m}{s^2}\).
In a situation where an object lays on a surface and does not leave the surface, the normal force is defined as follows:
The normal force is a force that occurs on an object with a magnitude such that the net force on the object in the direction perpendicular to the surface is zero.
As a result of this, the axes are generally defined so that the x-axis is parallel to the surface, and the y-axis is perpendicular to the surface. In some cases forces may need to be broken into components to find the magnitude of the particular force. This often happens when the object is on an incline.
Normal forces must exist due to a concept known as Newton’s Third Law. It states that any object which has a force exerted on it will unleash its own force of the same magnitude but the opposite direction (the fact that forces are vector quantities is extremely important for analyzing the existence of normal forces). Therefore, if a net force is applied on an object towards the surface it is on, it will “return” its own force of the same magnitude, giving the object a net force of 0 on that axis (typically the y-axis).
Now that it is clear why normal forces exist, let’s move onto some quantitative examples.
Normal force problems are broken into two categories: situations where the object is on a level surface, and situations where the object is on an incline. They are approached separately because the necessary problem-solving techniques for the two types of problems are rather different.
Example 2: A block laying on a flat table has a mass of 6 kilograms. A downward force of 12 Newtons is applied to the block. What is the normal force applied to the block?
Solution: Both the applied force and the force of gravity are downward forces, towards the center of the Earth. Therefore the normal force must be pointed upward. For starters, let’s find the force of gravity on the block:
$$F_g = mg = 6(9.8) = 58.8 \; N$$
The applied force is \(12\) Newtons of force in the same direction. Therefore, the total magnitude of the two downward forces is \(58.8 + 12 = 70.8 \; N\). The net force on the object is zero Newtons, so the normal force has the same magnitude as these combined forces but the opposite direction. Therefore the normal force is pointed upward and has a magnitude of \(70.8 \; N\).
Example 3: A block of mass 8 kilograms is on a flat, level, frictionless surface, and it is being pushed horizontally with a force of 12 Newtons. What is the normal force on the block?
Solution: According to Newton's Law, the horizontal acceleration can be found:
$$F_x = ma \Rightarrow$$ $$a = \frac{F_x}{m} = \frac{12 \; N}{8 \; kg} = 1.5 \frac{m}{s^2}$$
Since it is accelerating, it must be moving, so the normal force ensures that there is no movement only in the u-direction. Thus its magnitude will be the same as that of the force of gravity on the block. Since the force of gravity points downward, the normal force will point upward. Now find the magnitude:
$$|F_N| = |F_g| = mg = 8 \cdot 9.8 = 78. 4 \; N$$
However, if applied forces are not pointed parallel or perpendicular to the force of gravity, decomposition of the force vector into components will be necessary, as demonstrated in the next example.
Example 4: A block of mass \(2\) kilograms is on a frictionless surface. A force is applied to the block with a magnitude of \(12\) Newtons and is applied at 52 degrees below the horizontal. What is the normal force acting on the object?
Solution: Begin by listing all the forces acting on the object on each coordinate axis:
X-axis: X-component of the applied force
Y-axis: Y-component of the applied force Force of gravity Normal force
The object is not moving in the y-direction at all, so
$$F_N - Fy_{app} - F_g = 0$$
Solve for the normal force:
$$F_N = Fy_{app} + F_g$$
Now we can find \(Fy_{app}\) and \(F_g\):
$$\sin 52^{\circ} = \frac{Fy_{app}}{F_{app}} \Rightarrow$$ $$F_{app}\sin 52^{\circ} = F_y{app} \Rightarrow$$ $$F_y{app} = 9.456 \; N$$
Now find \(F_g\):
$$F_g = mg = 2 \cdot 9.8 = 19.6 \; N$$
The sum of these gives the magnitude of the normal force, \(29.056 \; N\).
Here is a different type of example that requires you to think backwards.
Example 5: A block is at rest on a flat, level surface. There are no forces acting on the block along the horizontal axis. The normal force on the block has a magnitude of \(124.9\) Newtons. What is the mass of the block?
Solution: The surface is level. Since the normal force always acts perpendicular to the surface, it points upward, opposing gravity. Due to the lack of other outside forces besides gravity, the normal force and the force of gravity have the same magnitudes:
$$F_N = F_g \Rightarrow$$ $$124.9 \; N = mg$$
We can solve for the mass of the block:
$$m = \frac{124.9 \; N}{g} = \frac{124.9 \; N}{9.8 \frac{m}{s^2}} = 12.745 \; kg$$
Now we investigate calculations of the normal forces of blocks that are on inclines. These are trickier because they require breaking up forces into components. As a rule of thumb, when the block (or other object) is on an incline, define the coordinate axes so that the x-axis is parallel to the surface and the y-axis is perpendicular to the surface (with this configuration the normal force will always be parallel to the y-axis). In general, the force of gravity will need to be broken into components.
Example 6: A block of mass \(8.5\) kilograms is set on a frictionless incline with an inclination angle of 20 degrees. What is the magnitude of the normal force on the block?
Solution: The force of gravity is not along either axis--it must be broken into components that are parallel to the two axes. The y-component opposes the normal force, and the x-component is parallel to the surface. So, according to Newton's Second Law:
$$\sum{F_y} = F_N - F_{gy} = ma_y$$
However the acceleration of the block on the y-axis is zero because there is no movement in that direction.
$$F_N - F_{gy} = 0 \Rightarrow$$ $$F_N = F_{gy}$$
However, \(F_{gy}\) needs to be written in terms of \(F_g\). Drawing a triangle with the legs as the components and the hypotenuse as the total gravitational force and using some basic triangle trigonometry gives
$$F_{gy} = F_g\cos 20^{\circ}$$
Thus,
$$F_N = F_g\cos 20^{\circ} = mg\cos 20^{\circ} = (8.5)(9.8)\cos 20^{\circ} = 78.276 \; N$$
We can use Example 6 as a conceptual model for handling similar problems, such as the final two examples, Examples 7 and 8.
Example 7: A block is on an incline with an inclination angle of 18 degrees. The normal force on the block has a magnitude of \(112.3\) Newtons. What is the mass of the block?
Solution: As with Example 6,
$$F_N = F_{gy}$$
With the same reasoning in Example 6, if we let the angle of incline be denoted as \(\theta\) (in degrees), then
$$F_{gy} = F_g\cos \theta$$
Therefore:
$$F_N = F_g\cos \theta \Rightarrow$$ $$F_N = mg\cos \theta$$
We have all of the variables' values except for the mass, which is what we need to solve for:
$$m = \frac{F_N}{g\cos \theta} = \frac{112.3}{9.8 \cos 18^{\circ}} = 12.049 \; kg$$
In general, for these types of problems, the angle of incline is denoted as \(\theta\).
Example 8: A block of mass \(4\) kilograms is on an incline. The normal force on the block has a magnitude of \(36.05\) Newtons. What is the angle of inclination?
Solution: In this type of problem, we know all of the relevant information except for \(\theta\), which makes setting up the equation a little bit harder. But, as is true in general:
$$F_{gy} = F_g\cos \theta$$
This means that
$$F_N = F_g\cos \theta \Rightarrow$$ $$F_N = mg\cos \theta$$
Now we need to solve for the angle. Isolate it:
$$\cos \theta = \frac{F_N}{mg} \Rightarrow$$ $$\theta = \arccos (\frac{F_N}{mg})$$
Now plug in the normal force's magnitude, the mass of the block, and the gravity constant:
$$\theta = \arccos (\frac{36.05}{4 \cdot 9.8}) = 23.126^{\circ}$$
Now it can be seen why these problems are broken into two categories: one category of problems are best handled on tilted axes by breaking the force of gravity into components. However, both types of problems demonstrate the importance of the normal force--none of the calculations in this section would be true without it, and you would not be sitting down reading this article without a normal force keeping you in your seat!