friction completed

content/physics/Forces.md

@@ -13,6 +13,8 @@ date: 2024-09-07
 - $\vec{F}$ is a force vector. Has a reference frame.
 - $\sum{\vec{F}}$ is the sum of all force vectors acting on an object. Also has a reference frame.
 - $m$ is mass, a scalar quantity
+- $\vec{F}_{fr}$ is taken to mean the force of friction.
+- $\mu_s$ is taken to mean the coefficient of static friction while $\mu_k$ is the coefficient of kinetic friction for that surface.
 # Newton's laws of Motion
 
 ## Force 
@@ -76,5 +78,42 @@ This is mathematically put down as $\vec{F}_1=-\vec{F}_2$. Below, $\vec{F}_{GP}=
 In the following illustration, Honest Abe exerts the force of gravity upon the table. Let's say he was about $0.9074074074$ Elons, or 1 Trump($245$ pounds). Firstly, illustration depicts 2 forces that art on the same object, Honest Abe. The downward arrow labeled $\vec{F}_G$ represents the force caused by gravity, and $\vec{F}_N$ represents the reactionary force enacted by the table upon the statue, meaning both act on the statue. A common misconception is that they're the same equal and opposite forces shown in Newton's 3rd Law -- because the 3rd law states that the reactionary force applies on a *different object.* The second image correctly displays the reactionary force to $\vec{F}_N$ as $\vec{F'}_N$, which is exerted on the table by the statue. **Don't ask what happens as a result of the gravitational force, Newton doesn't have a very good answer.**
 ![](gravity_forces.png)
 
+# Friction
+
+In Newtonian mechanics, Friction is the opposite *but not necessarily equal* force acting on the very object that tries sliding against a static, rigid body. Perfectly smooth objects do not impede moving objects through friction because friction is a result of the microscopic bumps normally found in any surface. When an object slides along a rough surface, the force of **kinetic friction** acts opposite to the direction of the object's velocity.
+> [!IMPORTANT]
+> Kinetic friction is not a result of Newton's $3^{\mathrm{rd}}$ law of motion.
+
+The force of friction was not derived through theoretical physics, but rather experimental results, so beware of MCQ choices that state the properties of friction absolutely. Through such empirical results, it was found that the force of friction was roughly proportional to the Normal force experienced by the object. The force of friction between hard surfaces is negligibly variant on the surface area in contact, so we can still work with Newtonian physics instead of dynamics. 
+
+Sliding friction is usually called **Kinetic Friction**, the magnitude of which force can be found using:
+$$
+||\vec{F}_{fr}||=\mu_k*||\vec{F}_N||
+$$
+
+There also exists **static friction** which refers to a force parallel to the two surfaces that can arise even when they are not sliding. Usually, this is the force that is exerted when trying to get an object to move. 
+$$
+0\le||\vec{F}_{fr}||\le\mu_s||\vec{F}_N||
+$$
+
+According to this formula, the force of static friction scales up until the object starts sliding along the surface, at which time the force of friction is now of the kinetic kind. You may also be provided tables like these:
+![](coef_friction.png)
+
+> [!NOTE] Example
+>   *Friction: static and kinetic.* A $10.0$ kg mystery box rests on a horizontal floor. The coefficient of static friction is $\mu_s=0.40$ and the coefficient of kinetic friction is $\mu_k=0.30$ Determine the force of friction $\vec{F}_{fr}$, acting on the box if a horizontal external applied force $\vec{F}_A$ is exerted on it of a magnitude: \
+>   A)  0 \
+>   B)  10N \
+>   C)  20N \
+>   D)  38N \
+>   E) 40N 
+
+> [!faq]- Answer
+>   $||\vec{F}_N||=9.8*10.0=98\mathrm{N}$, so $0\le||\vec{F}_{fr}||\le\mu_s*98$, where $\mu_s*98=0.4*98=39.2\mathrm{N}$ \
+>   A) static friction, $\mu_s=0$ \
+>   B) static friction, $\mu_s=10N$ \
+>   C) static friction, $20$N is still not sufficient to move the box \
+>   D) static friction, $38$N is still not sufficient to move the box, but just barley \
+>   E) kinetic friction, $40\mathrm{N}>39.2\mathrm{N}$, $||\vec{F}_{fr}||=0.30*98\mathrm{N}=29.4\mathrm{N}$. $40\mathrm{N}-29.4\mathrm{N}=10.6\mathrm{N}$, to see how fast the box will move: $\frac{||\vec{F}_f||}{10.0\mathrm{kg}}=1.06$ m/s. In freedom units: $2.371152$ Miles an hour.
+
 
 #physics