--- ## Definition - In physics a force is anything that **pushes or pulls an object on an object in some direction**. - A force requires the two objects/things that are involved to *interact* in some way. - The person/thing that is causing the force is usually called the **actor** and the object/thing that is receiving or experiencing the force is usually called the **recipient.** --- ## The basics - Force is measured in **newtons** which is denoted using the capital letter $N$. One newton is equal to the force required to accelerate a **1kg object to 1$\large\frac{m}{s}$**. Therefore, you should always convert mass into *kilograms* and distances into *meters* when using equations involving force $F$. - For context, a 5 pound weight exerts a force with a magnitude of 5N down on your hand. *here on earth at least.* - The **net force, denoted as $F_{\text{net}}$** is the summation of all the *individual* force vectors acting on an object. Since force is a vector quantity direction matters immensely, for example, the sum of two opposing force vectors is *not 2F* it is **zero**. - The net force should *(in magnitude)* roughly match the acceleration of an object at all times. If there is no net force acting on an object the acceleration *should be zero.* If there is a net force *the acceleration should be roughly equivalent in magnitude (relative to the mass of course).* >[!tip] >##### *Equilibrium* >- An object that is at rest **and not moving, (meaning it has zero velocity) is in a state of *static equilibrium***. >- An object that is moving with a **constant velocity, (meaning no acceleration) is in a state of *dynamic equilibrium***. >- In both of these states the object's **net force AND net torque *(see [[Rotational and Circular motion]])* must be equal to zero!** > - Therefore, if you are every asked a question like: how much of x force is needed to make x object be in a state of equilibrium? You know that the question is **really asking** how much of x force is needed to make the *net force* equal to zero *(and or net torque)*? Or another way of looking at it, how much of x force is needed to make the object's acceleration zero? > >$\newline$ >$\text{Static equilibrium: }\;\color{Bisque}\sum F_x = 0\;\;\;\sum F_y = 0\;\;\;\sum\tau = 0$ ##### *Newton's laws* >[!quote] > ###### **Newton's first law:** > *An object at rest will remain at rest and an object in motion will remain in motion until acted on by a force.* >- The law is the fundamental basis for the concept of *inertia*, the idea that objects want to continue being in the state of motion that they are currently in. If you want to change the motion of an object *(the velocity)* you must get that object to either accelerate or decelerate and this law proposes that acceleration is the **direct result of a force.** Without a force an object **cant accelerate or decelerate**. >--- >###### **Newton's second law** >*The acceleration of an object is directly proportional to the amount of force imparted on it and inversely proportional to its mass.* >- Newton's second law relates force to the acceleration and mass of an object. According to this law, doubling the force will double the recipient's acceleration, and doubling an object's mass will half its acceleration, *assuming the force imparted on it is constant.* > $\large F=ma\space\space\space\space\space\space\space\space\space\space\space\vec{F}=m\vec{a}$ > >--- >##### *Newton's third law* >*Newton's third law states that every action, or force has an equal AND opposite reaction.* >- This means that if an object A imparts a force onto object B, that object will **always exert a force of equal magnitude back onto object A in the opposite direction**. This is the reason why you cant *push* your hand through a wall. Every time you exert a force onto the wall with your hand the wall pushes *back with the same amount of force, thus keeping your hand in place (unit the wall breaks...)*. > - These forces are known as an **action and reaction pair**. > - Since the action/reaction forces are **acting upon *different* objects the net force felt by both objects is not zero**. > >![[forcereactionbetter2.png|850]] #### *Force vectors* - Forces **are vector quantities** since they have a magnitude *and* a direction. Therefore we can **visually represent forces** using arrows. The length of the arrow corresponds with the *magnitude* of the force and its orientation corresponds with the the *forces direction*. *(See more about vectors and physics in [[Free fall motion]]*)* - Since forces are vector quantities physicists often denote them using **vector notation**. Sometimes each force is given its own corresponding letter like $\vec{w}$ or $\vec{T}$ but most of the time we use $\vec{F}$ with a *subscript* to denote the **type of force**, $\vec{F_{s}}$ or $\vec{F_e}$. - If there is **more then one force** acting upon an object we can find the *direction* and *magnitude* of the **net force** by adding up all the contributing force vectors. - Just like with an object's velocity and acceleration vectors *(discussed in [[Free fall motion]])* physicists often split force vectors up into their individual components, **x and y** when performing calculations. You can find these two component vectors from a "combined" force vector using the following equations: - *Note: The variable $F$ without an arrow atop is used to denote magnitude ONLY. The variable $\vec{F}$, in vector form is used to denote both the force's magnitude **and** direction*. $\large\vec{F_x}=F\cdot \cos\theta\space\space\space\space\space\space\space\space\space\vec{F_y}=F\cdot \sin\theta$ >[!quote] >![[Forcesvectors.png|850]] #### *Free body diagrams* - In order to better understand how many different forces effect one object we draw something known as a ==**free body diagram**. These are diagrams of an object or a system that **visually show all relevant interacting force vectors**.== For example the free body diagram of a box on a table would be a drawing of a box on a surface with a vector labeled $\vec{w}$ pointing down and a vector labeled $\vec{n}$ pointing up. **Since we try to draw each vector to scale *(relative to the other vectors)* these two vectors should be roughly equal in magnitude.** - Using a free body diagram helps us better understand the direction and magnitude of the net force acting on an object because Its fairly easy to *visually* add up all relevant force vectors. The *estimated* net force vector is usually indicated somewhere on a free body diagram. - If the subject of your free body diagram is on an **incline** its important to **visually show that as if we were viewing it while parallel to the horizontal**. This means that the vector for the weight force should always be pointed directly down from *our perspective*. If you want *you can still rotate the x and y axis* to be in line with the slope *(which makes calculations much easier)* as long as that's indicated on the FBD. An example of this can be seen below: >[!quote] >##### *Free body diagram* >![[FBDbetter.png|850]] >[!caution] >##### *Random FBD tips:* >- When drawing the free body diagram of an ascending elevator coming to a stop the tension vector should be shorter than the weight vector to indicate the deceleration occurring. >- The normal force vector of an object on an incline should **always be shorter than the weight force vector**. >- If multiple objects are stacked on top of the subject of your FBD **we represent *each* of their weight forces as a separate, downwards facing vector**. Don't combine them into a single weight force vector. >- If a large block is pushing a smaller block that's in front of it the force vector of the large block *on* the small block should be smaller in magnitude then the force vector experienced by the actor onto itself. #### *Types of forces* - ==A **contact force** is any force that occurs as a result of two objects making contact with each other.== For example, when two objects collide they recoil backwards due to the contact force they imparted on the other. - Conversely, ==a **long range force** is a force that allows two objects to interact with each other even if they aren't in direct contact.== The strength of a long range force **is always proportional to the square of the distance between the interacting bodies**. >[!quote] >##### *Force notation guide:* >| *General* | *Spring* | *Weight* | *Tension* | *Normal* | *Static friction* | *Kinetic friction* | *Drag* | *Trust* | >|-|-|-|-|-|-|-|-|-| >| $\Large\vec{F}$ | $\Large\vec{F_{sp}}$ | $\Large\vec{w}$ | $\Large\vec{T}$ | $\Large\vec{n}$ | $\Large\vec{f_s}$ | $\Large\vec{f_k}$ | $\Large\vec{D}$ | $\Large\vec{F_{\small\textit{trust}}}$ | --- ## Weight force - ==**The weight force $\vec{w}$** is the force imparted by gravity onto an object.== If an object weighs more, gravity pulls on it more strongly which is why the weight force experienced by an object is correlated directly to its **mass**. The weight force is also effected by the **strength of gravity**, for example, since the moon has less gravity then earth, objects there feel lighter *(Less weight force is acting on them)*. - The weight force **always points directly downwards regardless of the objects orientation** and is *usually* omnipresent in all situations. - An objects ==**apparent weight is the equal to the magnitude of the *contact forces* that prevent the object from falling through the ground**.== Without these contact forces, you would be, from your perspective weightless *(think falling in a 0 g plane)*! Since the "weight" of an object is solely due to its *relative frame of motion* we have to make a distinction between the apparent and "true" weight *(depends on where the observer is)*. $\text{Equation for weight: }\space\color{LightCyan}\large \big(w=mg\big)$ $\text{Equation for apparent weight: }\space\color{LightCyan}\large\big(w_{\text{app}}=w\space +ma_y\big)\color{LightGrey}=n$ $\newline$ - When an object's *relative frame of motion* is accelerating upwards, opposite the direction of gravity the *the object's apparent weight is greater than its actual weight.* However, when the relative frame of motion is traveling down the object's *apparent weight is less than its actual weight*. - To find the downwards acceleration needed for an *object's apparent weight to be 0* we can use the above equation. The result of this calculation is -9.8 $\frac{m}{s^2}$ which makes sense because it's **equal** to the acceleration due to gravity. $\text{Weightless when: }\space w_{\text{app}}=0\longrightarrow\frac{-\cancel{m}g}{\cancel{m}}=-g=\color{LightCyan}-9.8\frac{m}{s^2}$ $\newline$ >[!danger] >##### *The difference between mass and weight* >- Mass describes an object's *inertia* - that is, the degree to which it resists acceleration or deceleration due to a force. >- Weight on the other hand, **is a force,** specifically its the force that **gravity** exerts onto an object, pulling it downwards. The weight force experienced by an object is however, **directly proportional to the object's mass**. >##### *Apparent weight* >- The magnitude of the weight force is **distinct from an object's *apparent weight*** which is proportional **to the normal force exerted on that object**. --- ## Spring force - ==**The spring force $\vec{F_{sp}}$** Is the force imparted on an object by a *spring*==. This force is also sometimes called the *restorative force* since it can acting in either direction, doing whatever it can to restore equilibrium. If you **compress** a spring it will intact a force *towards you*, resisting compression. On the other hand if you pull on a string it will intact an inward *compressing force* attempting to return back to its resting state. - For more information on the spring force see [[Springs and elasticity]]. --- ## Tension force - ==**The tension force $\vec{T}$** is the force imparted on an object when it is **pulled on by a string**.== The direction of this force is **always in the direction of the spring** since objects cant be pushed with a string, only pulled. On a microscopic level, the tension force is due to the **molecular bonds** of the material that makes up the rope. When the rope is tugged, the bonds, like magnets try to keep the material together by pulling themselves back together. - The tension in an **ideal spring is the same at all points along that string**. *Note that an ideal string is a string that has no mass*. - To find the **tension** in a string we must **add up** the forces that are pulling that **string from either side**. *Make sure you **add** these forces even if they are acting in different directions*. - If a force is being applied to a rope **at an angle thus, causing that rope to stretch diagonally** we need to calculate the x and y components of the rope's tension separately and then combine them to find the final tension on that rope. ##### *Tension calculator* <iframe src="https://amesweb.info/Physics/Free-Fall-Calculator.aspx" width ="100%" height="600"></iframe> --- ## Normal force - ==**The normal force $\vec{n}$** is the force that a surface imparts on an object **that prevents the object from falling or crumbling into the surface**.== For example, when a book is placed on a table the reason it doesn't fall through the table is because the table exerts a normal force directly back up into the book. This is due to the molecular bonds in the wooden tabletop resisting compression. - The direction of the normal force exerted on an object **is always directly perpendicular to the surface that the object is on**. Furthermore, the magnitude of the normal force **is always equal in magnitude to the component of the weight force directly opposite it**. This is an important distinction to make because when an object is on an incline the normal force will be less than the overall weight force since it is offset from it by an angle. *See image below.* - The magnitude of the **normal force exerted on an object is proportional to the *apparent weight* of that object**. The sensation of weight is due to the normal force exerted by a surface onto and object *(or a person!)*. Therefore we say that an object's apparent weight is equal in magnitude to the normal force. For example, the reason why you feel weightless in a falling plane is not because your weight force **has changed** but rather because the floor of the plane is *also* falling and therefore cant exert a normal force onto you. >[!quote] >![[Forcesweightforce.png|850]] >[!caution] >##### *The normal force and inclines* >- The normal force exerted onto and object that is on an incline is equal to the **component of the weight force that is perpendicular to the surface of the incline**. > - This is usually the *y* component of the weight force *(assuming the coordinate grid is aligned with the incline)*. To find this component use $\cos(\theta)$ rather than $\sin$ because the angle of the incline to the horizontal corresponds with a different angle in the "triangle" made from the weight force and it's components. >- The normal force will be **less than the weight force** only when an object is on **an incline**. If the normal force were equal, you wouldn't need to add some additional, *upwards force* to stopping the object from moving/rolling down the incline! --- ## Friction force - ==**The friction force $\vec{f}$** is the force imparted onto an object that is moving along a surface or other object.== The friction force is always in the opposite direction of the object's velocity since it is **resisting that motion**, trying to slow the object down. Friction occurs because at a microscopic level every object/surface is somewhat rough, moving along this rough surface requires more force since the objects scrape against each other. - The friction force experienced by an object is **not dependent on the surface area of that object or the velocity of the object!** >[!caution] >- Friction can be in the same direction as the direction of motion **if it is keeping an object pinned to some moving surface.** For example, think of a cup sitting on a cart that is accelerating to the right. So long as that cup does not **slip of the back of the cart**, the force that is keeping it *"pinned"* to the cart and moving in tandem is friction. - ==**Static friction** denoted as **$\vec{f_s}$** acts on stationary objects.== It is the force that prevents heavy objects from moving when you tap on them. **The point at which the static friction force *is no longer strong enough to keep an object stationary,*** is known as the static friction maximum or $f_{\text{max}}$. It is calculated by multiplying the *coefficient of static friction (a number that depends on the object's material)* by the *normal force* experienced by the object. - ==**Kinetic friction** denoted as **$\vec{f_k}$** acts on objects *sliding along a surface*.== It is the force that opposes the motion of a **moving object**. Just like with static friction the magnitude of the **kinetic friction force depends on the *coefficient of kinetic friction*** which is dependent upon the object's material. Also, like static friction the *greater* the magnitude of the normal force *the greater* the magnitude of the frictional force which makes sense because heavier objects are harder to push/pull. - The coefficient of kinetic friction is always less than the coefficient of static friction for the same two materials. $\large f_{s}=\mu_sn\space\space\space\space\space\space\space\space\space\space f_{k}=\mu_kn$ $\small\text{Recall on a flat surface: }n=w=mg$ $\newline$ - When you are pushing on a a heavy object and **it is not moving** that means that the **static friction force is equal to the pushing force**. Once the object *starts to move* that indicates that the force pushing on the object *has overcome the static friction force* and the object is now experiencing a **kinetic friction force**. The kinetic friction force **is not equal in magnitude to the pushing force**, if it was the object would not be moving. $\text{When stationary: }\big(F_{\text{push}}=f_s\big)\space\space\space\space\space\space\space\space\text{When velocity is constant: }\Big(F_{\text{push}}=f_k\Big)$ $\text{When accelerating: }\big(F_{\text{push}}\ge f_k\big)\space\space\space\space\space\space\space\space\text{When decelerating: }\Big(F_{\text{push}}\le f_k\Big)$ $\newline$ >[!tip] >##### *Useful friction facts* >- When friction is involved objects that are being pushed or pulled by a force **don't have to be accelerating.** If the pushing force ceases than the object will *rapidly decelerate* coming to a stop since the friction force **always opposes motion**. >- The friction force felt by an object is **directly proportional to the weight of that object**. In other words the harder an object is pressed up against a surface, the more friction it will experience, *(common sense really)*! >- The strength of the friction force **is directly proportional to the weight of the object**. Or more accurately, the magnitude of the normal force. >- Usually physicists say that friction is **a force exerted *ON* the object**, meaning it comes from an outside source, the surface! >- Friction force magnitude does not depend on speed, and it is proportional to the normal force exerted onto an object. --- ## Drag force - ==**Drag or air resistance** is the force imparted on an object that is traveling through some medium *(gas, water...)*.== It exists because an object that is traveling through a medium must *push* molecules out of its way. The direction of the **drag force exerted on an object is always opposite to its direction of motion**, and the magnitude of this drag is always proportional to the object's velocity. - Since drag can be the result of either inertial forces *(medium resists object's movement because it is large/moving fast and causing the medium to move from rest)* or viscous forces *(the medium resisting the motion of an object because of thickness, "grabbing" the object)* there are ***two different ways we calculate drag.*** To quantify weather the drag experienced by an object is due to viscous or inertial forces we can calculate an object's Reynold's number. The magnitude of this number determines which equation for **drag force should be used.** - *In the equation for Reynold's number $\rho$ is the density of the medium/fluid, $v$ is the velocity of the object, $L$ is a measure of the object's size and $\eta$ is the viscosity of the fluid.* $\newline$ $\text{Reynold's number: }Re=\frac{\small\textit{inertial forces}}{\small\textit{viscous forces}}=\large\color{LightCyan}\frac{\rho vL}{\eta}$ $\newline$ - Drag at high Reynold's numbers is known as **quadratic drag** because the drag increases **exponentially with velocity**, *denoted by the square $v$ term.* Calculating the magnitude of drag due to inertial forces like this requires **the constant $C_D$ which is known as the *drag coefficient*.** This is a value which depends on the *shape* of an object *(low $C_D$ shapes are more aerodynamic)*. - Drag at **low Reynold's numbers is known as *linear drag*** because the magnitude of the drag for is **proportional** to the object's speed. When the Reynold's number is very small the object is also *usually* very small, therefore we generalize its shape as a sphere with **radius $r$** *(this equation is known as Stroke's law)*. In viscous mediums the **shape of an object has less effect on the magnitude of the drag force** it experiences. - The equations below give you the **magnitude of the drag force** for objects/situations with both *low* and *high* Reynold's numbers. $\newline$ $\text{Drag force at high Reynolds number: }\Big(Re\ge1000\Big)\longrightarrow\space\color{LightCyan}\large D=\biggl(\frac{1}{2}C_D\cdot\rho\cdot A\cdot v^2\biggl)$ $\text{Drag force at low Reynold's number: }\Big(Re<1\Big)\longrightarrow\color{LightCyan}\large D=\Big(6\pi\cdot \eta\cdot r\cdot v\Big)$ $\newline$ >[!tip] >##### *Terminal velocity* >- Terminal velocity is the speed at which the drag force exerted on an object **is equal to the weight force**. Since both the drag and weight force act in **opposite directions** they will eventually cancel each other out leading to a net force of 0. When the net force is 0 the object will stop accelerating and the speed at which this happens is the object's *terminal velocity*. > - We can calculate the terminal velocity of an object because we know that the **drag force of an object at terminal velocity will be *equal* to its weight force**. Then we can re-arrange the equation for drag at high Reynolds numbers to solve for v. > >$\textit{At terminal velocity: }|\vec{D}|=|\vec{w}|=mg$ >$\therefore\; |\vec{D}|=\frac{1}{2}C_D\rho Av^2=mg\;\longrightarrow\;\color{Bisque}v=\sqrt{\frac{2mg}{C_D\cdot\rho\cdot A}}=\textit{terminal velocity}$ >- When the Reynold's number is extremely low, terminal velocity is reached **almost instantly** since viscous liquids resist motion even when velocity is very low. Conversely when the Reynold's number is very high, the object's terminal velocity will also be proportionally large. >##### *Apparent weight and drag* >- Since drag increases as an object's velocity increases the **apparent weight** that the object feels increases until terminal velocity has been reached. *NOTE: In this case the drag force serves the same purpose as the normal force, which usually dictates apparent weight.* >##### *Life at low Reynold's numbers* >- Swimming through water is **relatively easy** because it isn't very viscous which allows inertia to play a role. For example, when a fish moves its fin it pushes water away from it which generates **thrust, pushing the fish in that direction**. When the fish moves its fin in the opposite direction, **inertia prevents it from *instantaneously* changing direction, which would have negated its prior movement.** *Recall, inertia makes it difficult for objects in motion to stop being in motion.* The fish can continue doing this over and over again to move through the water. >- Swimming through a very **viscous fluid is extremely hard** because object's reach their terminal velocity almoast instantly, negating the effect of inertia. If a fish were to move its find *back and forth* it would just move *back and forth*. Without the effects of inertia objects that are moving can **instantly come to a stop and move in the other direction**, in other words its impossible to "coast". > - Since water is incredibly **viscous on the microscopic scale** small single celled organisms and bacteria use *propeller like* appendages to apply a **constant force in a constant direction**. This method of movement does not rely on the effects of inertia like the back and forth motion of a fin. > >![[Forcessmallreynoldsnumbermotion.png|850]] --- ## Trust force - ==**Thrust** is a **reaction force** described quantitatively by **Newton's third law**. When a system expels or accelerates a **mass** in a direction the accelerated mass will **cause a force of equal magnitude but in the *opposite direction*==. - Traditionally **thrust** is generated by converting [[Energy|chemical energy]] to kinetic energy through **combustion**. - Trust is the result of **newton's third law of motion**. "*Every action has an equal and opposite reaction.*" - Thrust is generated due to the law of **[[Momentum|conservation of momentum]] which states that the *"amount of momentum in a system must remain constant"***. When an object with mass is **accelerated** in a direction its **momentum changes**. Since momentum is conserved within a system this elicits an equal **and opposite change in the momentum of the thrust generating object**, thus causing it to accelerate in the **opposite direction**. - We can **calculate the change in force *due* to the change in momentum** by using the following equation where $\small m$ is an object's *mass*, $\small v$ is the *velocity* and $\small t$ is the *time*. $\text{Force due to momentum: }\;\;\Huge\boxed{\large\color{lightcyan} F=\frac{[mv]_{\small 1}-[mv]_{\small 2}}{t_{\small f}-t_{\small i}}}$ --- ## Force and motion - When a physicist is attempting to predict, mathematically how a certain force will effect the *motion* of an object they will usually split both the force and motion into their individual **x and y components**. This makes the process of calculation easier since we can **individually calculate how both the x and y component of an object's motion changes as a result of the x and y force components acting on it.** - We can also use Newton's second law to calculate how the forces imparted on an object effect its **acceleration**. We do this by first calculating the **net force in both the x and y directions** by breaking each force up into their individual components *using $\sin(\theta)$ and $\cos(\theta)$* and adding them together using the equations shown below. - Once we know the x and y components of the *net force* we can calculate the acceleration those forces impart onto the object by using newton's second law, re-arranged in the following way: $\newline$ $\text{Newtons second law: }\;\;\Huge\boxed{\large\color{LightCyan}F=ma\longrightarrow a=\frac{F}{m}}$ $\newline$ - Once we know the acceleration of the object in the x and y directions we can use the kinematics equations outlined in [[Motion dot diagrams]] and [[Free fall motion]] to calculate how that changes the velocity and or position of the object. *If there is not net force in the x or y direction* recall that the object will **continue moving in that direction at constant velocity**. If we are to change the motion of an object in any way *(acceleration or deceleration)* there must be a net force *(by newton's second law).* $\newline$ $\large\text{Newton's second law in component form:}$ $\newline$ ${\large F_{x\space\text{net}}=\color{LightCyan}\sum\big(\text{\color{grey}\small x component of all forces}\big)=ma_x}\color{LightGrey}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{\small\color{grey}for example: }\space\sum\space n_x+f_x+w_x=ma_x$ ${\large F_{y\space\text{net}}=\color{LightCyan}\sum\big(\text{\color{grey}\small y component of all forces}\big)=ma_y}\color{LightGrey}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{\small\color{grey}for example: }\space\sum\space T_y+w_y+n_y=ma_y$ $\newline$ $\text{Other useful equations:}$ $\newline$ $\text{Friction and acceleration: }\large\space\color{LightCyan}\Big(-f=ma\Big)$ $\text{Weight and acceleration: }\space\space\color{LightCyan}\large \big(-w=-ma\big)$ $\newline$ >[!tip] >##### *Consequences of newton's second law* >- If a force of constant magnitude and direction is applied to an object, the object will move with a **constant acceleration** in the corresponding direction. The magnitude of this acceleration **is directly proportional to the magnitude of the force** imparted on the object. >- If a force of constant magnitude and direction is applied to two objects of differing mass, they will both experience a different amount of acceleration. This is because **mass and acceleration *(from a force)* are inversely proportional**. This is why it is *more difficult* to accelerate a massive object to the same velocity as a less massive object. ##### *Objects in contact* - If two objects are in *direct* contact with one another and the pair is being pushed in a direction by some external force, that force will be ***"split up" amongst the two objects relative to their mass***. In other words, the force experienced by each object **is not equal to the force that is exerted on the pair overall**. - To calculate the acceleration of this "pair" of objects **use their combined mass, pretending as if the objects have been merged into one.** Plug that mass into Newton's second law equation: ${\color{LightCyan}a=\frac{F_{\textit{overall}}}{m_{\textit{combined}}}}=\text{Acceleration of objects in contact}$ - The magnitude of the net force that *each* object feels **will be less then the overall force**. The object in direct contact with the actor or, the outside force *(pushing on the pair of objects)* will experience an external force denoted $\small\vec{F}$ and the object next to it will experience a **normal force** denoted as $\vec{n}$ by the other object. To calculate these two forces we can use the following formula: *where $F_1$ is the force felt by one of the objects in the pair and $m_1$ is the mass of that object.* ${\color{LightCyan}F_1=m_1\cdot a_{\textit{overall}}}\;\;\text{\color{grey}where }\;a_{\text{overall }}=\frac{F_{\textit{overall}}}{m_{\textit{combined}}}$ >[!tip] >##### *FBD of objects in contact* >- The block that the external force is acting on will have two **horizontal forces acting on it**. This first is the external force *on "block A"* which is directed towards the other block, *"block B"*. the magnitude of this force **will be the larger than *any other* horizontal force in the system because it must be "able" to *push both blocks.*** >- Block **A and block B will experience a normal force of equal magnitude in opposite directions** *(due to Newton's third law).* The magnitude of this normal force is equal to the force needed to push block B with the same acceleration as the combined system of blocks. > - The difference between the normal force **on block A** and the external force **on block A** ==will be equal to its net horizontal force. The magnitude of this net force will equal the force needed to push block A with the same acceleration as the overall system.== > - Block B will only experience **one** horizontal force: the normal force exerted on it by A. However, if there was another block next to it *(block C)* block B would experience a **"backwards" normal force just like block A.** Furthermore, ==the *difference* between the normal force exerted on *block B by A* and the normal force exerted on *block B by C* will be **equal in magnitude to the force needed to "push" block B with the same acceleration as the combined system.**== *(just like how it was for block A)* >##### *FBD of situation* > >![[Pasted image 20230502191759.png|340]]![[Pasted image 20230502191812.png|519]] > ##### *Ropes pullies and tension* - In physics ropes are usually assumed to be **massless, non-elastic objects and pullies/wheels** are assumed to have **no friction**, these are known colloquially as **ideal ropes/pullies/wheels.** Ideal ropes essentially, **transmit** force in one direction only and pullies/wheels change the direction of force *(pullies can be a bit more complex)*. - The **tension in an ideal rope is equal to the magnitude of the force pulling it** and the tension is the **same** at all points along the rope. If two force are acting on a rope from opposite sides the **overall tension in the rope will be equal in magnitude to *whichever force is lower***. This is because a ==rope pulls on the objects at each of its ends with a force equal in magnitude to the tension, which is equal in magnitude to the *lesser force*==. $\sum F_x=F-F_{\text{\small (object on rope})}=F-T=m_{\text{\small{(rope)}}}a_x=0$ - When a rope passes through an **ideal pully the magnitude of the tension in that rope *does not change*.** What does change however, is the direction of the force/tension. Therefore, when setting up calculations and FBDs for situations that involve pullies we can pretend like the force acting on one end of the rope is **identical to the force acting on the other end, no matter how the rope's direction has changed.** - If a rope has been strung through multiple pullies **in a chain** the force required to lift an object **will be reduced by the *reciprocal* of the number of pullies *in that chain***. *If there is just 2 pullies, the force required is half, 3 pullies a third, four pullies a fourth and so on.* However, with each new pully added to the chain, the ratio between the distance the object moves vs the distance the rope must be pulled increases exponentially. *2 pullies and the distance needed doubles, 3 pullies and the distance needed … and so on...* >[!example] >##### *Rope, pully and tension example* >![[Forcespullyquestion 2.png|850]] >- The tension in **both ropes is equal**. In rope 1 50kg of force is pulling on each end of the rope *(rope pulls on the wall which by Newton's third law, pulls on the pully)* and the same is true for pully 2, therefore they experience the *same tension.* > --- ## Calculator <iframe src="https://app.calconic.com/api/embed/calculator/648a452928c7470029d3eba4" sandbox="allow-same-origin allow-forms allow-scripts allow-top-navigation allow-popups-to-escape-sandbox allow-popups" title="Calconic_ Calculator" name="Calconic_ Calculator" height="12000" scrolling="no" style="width: 100%; border: 0; outline: none;"></iframe>