Friction and Drag (Mechanics)

#Mechanics # A Brief Introduction In [[Newton's Laws and SUVAT|our fundamentals document]], we look at Newton's Third Law. It is noted that there must be a reaction force if there is an action force causing a change in motion. In actuality, **friction** is this reaction force when an object moves across a solid surface while **drag** is the reaction force when an object moves through a fluid, like air. # Friction Note there are actually two forms of friction, static friction and dynamic friction. More information below: ## Static Friction This form of friction occurs when there is a resultant force on an object but no change in motion. This simply implies that there is a frictional force larger than the force opposing it. **The formula used to define static friction is:** $ f_{smax} = \mu_sF_n$Where: - $\mu_s$ is the coefficient of static friction - $F_n$ is the magnitude of the vertical normal force, $mg\sin \theta$. This value is a nominal maximum to when static friction can take place. The direct static friction component $f_smax$ is just equal to the magnitude of the force component acting on it. >[!Example]- EXAMPLE PROBLEM: Module 6.1 Problem 1 >The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.25 with the floor. If the train is initially moving at a speed of 48 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor? > >**Solution:** >It is given that the resultant force of the train will be constant, as the acceleration is constant and the train doesn't accelerate upwards. Therefore: >$ma = 0$ > >As there is no resultant force, the normal force on the crates **must** be equal to $-mg$, we can find the resultant friction as: >$\mu_{s}F_{N} = 0.25 x -9.81 = -2.45 ms^{-2}$ >This is our deceleration. Use the [[Newton's Laws and SUVAT#The SUVAT Equations|SUVAT Equations]] to get an expression for $s$! >$0 = 48 \times \frac{1000}{3600} - 2.45 t$ >$t = 19.6 ms^{-1}$ >$s = 48 (19.6) - \frac{1}{2}2.45 \times 19.6^2 $ >$s = 470.2 ms^{-2}$ >$s$ is our displacement - that's why you see those videos f train crashes! It really isn't easy to stop a moving hunk of steel... ## Dynamic Friction This form of friction is given as a constant $F_k$ if the body in question is sliding on the surface. The formula used to find the magnitude of the frictional force **when the object is moving** is: $f_k = \mu_kF_n$ Where $F_n$ is the magnitude of the vertical normal force, $mg\sin \theta$. This coefficient's going to be smaller than the static one - this is as the surface is pushed further away by a moving object than an object at rest. >[!EXAMPLE]- EXAMPLE PROBLEM: Module 6.1 Problem 2 >In a pickup game of dorm shuffleboard, students crazed by fi-nal exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.90 m by the horizontal 25 N force from the broom and then has a speed of 1.60 m/s, what is the coefficient of kinetic friction be-tween the book and floor? >We can find the resultant force of the system first: >$F = 3.5 \times (\frac{1.6}{0.9}) = 6.22$ >Therefore the magnitude of the kinetic frictional force $f_{k}$ is >$f_{k} = 25.0 - 6.22 = 18.78N$ >Giving us an expression for the coefficient of kinetic friction: >$\mu_{k} = \frac{18.78}{25} \approx 0.75 $ # Drag Drag is the normal force that acts against a change in motion of an object if the object is travelling through a fluid, such as air. [[Fluid Dynamics Basics (Mechanics)|Learn more about fluids with our introductory fluid dynamics' course! ]] The normal force $\overrightarrow{D}$ opposes the relative motion of an object and points in the direction opposite the fluid flow. We can use the following formula to express the drag force $\overrightarrow{D}$: $\overrightarrow{D} = -\frac{1}{2}C \rho Av^2$ Where: - $\rho$ is the density of the fluid medium, - $C$ is the **drag coefficient** of the body, a constant that can be found - $A$ is the relative **perpendicular** cross-sectional area to the direction of fluid flow - $v$ is the relative speed of the object against the fluid >[!EXAMPLE]- Example Problem: Module 6-2 Problem 39 >Calculate the ratio of the drag force on a jet flying at 1000 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at half that speed and altitude. The density of air is 0.38 kg/m3 at 10 km and 0.67 kg/m3 at 5.0 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C. >Substitute all possible values and create two different equations, each describing the motions at different altitudes: >$D_{1} = \frac{1}{2}0.38AC(1000)^2$ >$D_{2} = \frac{1}{2}0.67AC(1000)^2$ >Now we divide D1 by D2 to find the ratio: >$\frac{D_{1}}{D_{2}} = \frac{\frac{1}{2}C0.38A(1000)^2}{\frac{1}{2}C0.67A(1000)^2}$ >$\frac{D_{1}}{D_{2}} = \frac{190000}{335000}$ >Therefore the ratio between $D_{1}$ and $D_{2}$ is $38 : 67$ %%power due to drag force:%% $P = \frac{1}{2}C\rho Av^3$ As work done = force x distance (at term. vel given by drag force) so power = force x speed ( as $P = \frac{W}{t}$) this is the required power to overcome a force! explain how to take it with both forces and energies now… ## Terminal Speed People like to skydive, some a lot! Apparently, the record for the highest unaided (so no parachute!) skydive is around **41.4** kilometres! So at some point, given that there's a reaction force $\overrightarrow{D}$ acting against the weight of something falling $mg$, we'd expect both forces to be equal, right? This is called the **Terminal Velocity** of an object. It is given by the equation: $v_{t} = \sqrt{ \frac{2F_{g}}{C\rho A} }$ >[!Abstract]- Deriving the formula >We can express the acceleration of an object at terminal velocity as 0. >This means we can write the formula of the resultant as $\overrightarrow{D} - F_{g} = 0$ where $F_g$ is the downwards force due to gravity. >We can then expand and rearrange the formulae so that $v_t$ is isolated. >$\frac{1}{2}C\rho Av^2 - F_{g} = 0$ >$C\rho Av^2 = 2F_{g}$ >$v_t = \sqrt{\frac{2F_{g}}{C\rho A}}$ >[!Example]- Example Problem: Module 6-2 Problem 36 >The terminal speed of a sky diver is 160 km/h in the spread-eagle position and 310 km/h in the nosedive position. Assuming that the diver’s drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position. > >**Solution** >We can take the value of the terminal velocity $v$ as a function of the weight, cross-sectional area and the drag coefficient by using the terminal velocity function from above. >We can also see that only the cross-sectional areas change, allowing us to treat all the other variables as constants. We can thus create two equations: >$160 = \sqrt{ \frac{2F_{g}}{C\rho A_{1}} }$ >$310 = \sqrt{ \frac{2F_{g}}{C\rho A_{2}} } $ >Reorganise the function so that $A$ is the subject. >$\frac{25600C\rho}{2F_{g}} = \frac{1}{A_{1}}$ >$\frac{91600C\rho}{2F_{g}} = \frac{1}{A_{2}}$ >Divide the expression for $A_1$ by the expression for $A_2$. This should allow the constants to cancel. >$\frac{\frac{1}{A_{1}}}{\frac{1}{A_{2}}} = \frac{\frac{25600C\rho}{2F_{g}}}{\frac{91600C\rho}{2F_{g}}}$ >$\frac{\frac{1}{A_{1}}}{\frac{1}{A_{2}}} = \frac{25600}{91600}$ >$\frac{A_{1}}{A_{2}} = \frac{91600}{25600}$ >Therefore the ratio of the slower object's cross-sectional area to the faster object's area is $3.58:1$ >[!EXAMPLE]- Example Problem: Module 6-2 Problem 40 >In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is u = 40.0$^\circ$, the snow is dry snow with a coefficient of kinetic friction $\mu_k$ 0.0400, the mass of the skier and equipment is m = 85.0 kg, the cross-sectional area of the (tucked) skier is A = 1.30 $m^2$, the drag coefficient is C = 0.150, and the air density is 1.20 kg/$m^3$. (a) What is the terminal speed? (b) If a skier can vary C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed? > >**Solution a)** >Remember the equation for terminal speed given above. Just to reiterate, this is: >$v_{t} = \sqrt{ \frac{2F_{g}}{C\rho A} }$ >Substitute values! They're all given above. This yields us a value for the terminal speed: >$v_{t} = \sqrt{ \frac{2 \times 85 \times 9.8}{0.150 \times 1.20 \times 1.30 }} = 84.4 m / s$ > >**Solution b)** > >This is an example of a limits question. We can create an equation to gauge the sensitivity of the variable $v_t$ in relation to a minute change $dC$, which subtends to zero. (Note [[More on Limits (Maths)|L'Hopital's Rule]]) > >This means the answer that we get will be relative to dC. The equation we can construct will be as follows (using the small approximations rules found in the calculus booklet): >$dv_{t} = \sqrt{ \frac{2F_{g}}{dC\rho A}}$ >Plugging in numerical constants gives us an expression for the change in $v_t$ as follows: >$dv_{t} = \sqrt{ \frac{2 \times 85 \times 9.8 }{dC \times 1.20 \times 1.30}} = \frac{1068}{dC}$ >...please check if this is right # What's Next? For the next instalment in our mechanics topic, here: [[Uniform Circular Motion (Mechanics)]] To go back to the main physics page, here: [[The (Incomplete) Physics Almanac]]