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anonymous
 one year ago
A curve of radius 76 m is banked for a design speed of 100 km/h. If the coefficient of static friction is 0.38 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]
anonymous
 one year ago
A curve of radius 76 m is banked for a design speed of 100 km/h. If the coefficient of static friction is 0.38 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]

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rvc
 one year ago
Best ResponseYou've already chosen the best response.0r= 76m v= 100 km/h > 100 X 5/18 m/s \(\mu\)= 0.38

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@rvc it needs a max and a min /:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1on our car are acting two forces, namely: dw:1435475379190:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the condition for v_max, is then: \[\Large m\frac{{{v^2}}}{R} = \mu mg\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1which comes from the subsequent condition: \[\Large {F_{centrifugal}} = {F_{friction}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1solving the last expression for v, we get: \[\Large {v_{\max }} = \sqrt {\mu Rg} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1actually our curve is, like below: dw:1435475861915:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have this vector decomposition: dw:1435476024271:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we can write: \[m\frac{{{v^2}}}{R}\cos \theta = \mu mg\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large m\frac{{{v^2}}}{R}\cos \theta = \mu mg\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1using initial data, we get: \[\Large \cos \theta = \frac{{\mu gR}}{{{v^2}}} = \frac{{0.38 \times 9.81 \times 76}}{{{{\left( {100/3.6} \right)}^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435476363488:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435476471088:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1you are right! @rvc

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the right friction force is: \[\Large \mu \left( {mg\cos \theta + m\frac{{{v^2}}}{R}\sin \theta } \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so the right equation is: \[\large m\frac{{{v^2}}}{R}\cos \theta = m\mu \left( {g\cos \theta + \frac{{{v^2}}}{R}\sin \theta } \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops.. I forgot the component of the weight of the car along the inclined plane: \[\large m\frac{{{v^2}}}{R}\cos \theta = m\mu \left( {g\cos \theta + \frac{{{v^2}}}{R}\sin \theta } \right) + mg\sin \theta \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if we can neglect the weight of our car, then we can write: \[\Large \mu m\frac{{{v^2}}}{R}\sin \theta = m\frac{{{v^2}}}{R}\cos \theta \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1or: \[\Large \tan \theta = \frac{1}{\mu }\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the external forces acting on our car, are the subsequent: dw:1435478456892:dw where N is the perpendicular reaction of the plane, R is the friction force3, Fc I the centrifugal force, and W is the weight of our car. So we can write the subsequent vector equation: \[\Large {{\mathbf{F}}_{\mathbf{C}}} + {\mathbf{W}} + {\mathbf{N}} + {\mathbf{R}} = {\mathbf{0}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1which is equivalent to these 2 scalar equations: \[\Large \left\{ \begin{gathered} m\frac{{{v^2}}}{R}\cos \theta  mg\sin \theta  \mu N = 0 \hfill \\ \hfill \\  m\frac{{{v^2}}}{R}\sin \theta  mg\cos \theta + N = 0 \hfill \\ \end{gathered} \right.\] where N is the magnitude of the vector N

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1solving the second equation for N, and substituting into the first equation, we get: \[\large m\frac{{{v^2}}}{R}\cos \theta  mg\sin \theta  \mu \left( {mg\cos \theta + m\frac{{{v^2}}}{R}\sin \theta } \right) = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1after simplification, we get: \[\Large {v^2} = gr\frac{{\mu + \tan \theta }}{{1  \mu \tan \theta }}\] where r is the radius of our curve

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we get the min value of v, if we set: \[\Large \tan \theta = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1better is if we study this function, using the mathematical analysis: \[\Large f\left( \theta \right) = \frac{{\mu + \tan \theta }}{{1  \mu \tan \theta }}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got that function, is an increasing function. Nevertheless I want to check my conjecture using this one: \[\Large {v^2} = gr\frac{{\mu \cos \theta + \sin \theta }}{{\cos \theta  \mu \sin \theta }}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1which is equivalent

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got this: \[\Large \frac{d}{{d\theta }}\left( {\frac{{\mu \cos \theta + \sin \theta }}{{\cos \theta  \mu \sin \theta }}} \right) = \frac{{1 + {\mu ^2}}}{{{{\left( {\cos \theta  \mu \sin \theta } \right)}^2}}}\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.0well now let the user find out the equation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! I can state this, since that function is an increasing function, then the min value is at \theta=0, namely: \[\Large \tan \theta = 0\] and therefore: \[\Large {v^2} = \mu gr\]
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