anonymous
  • anonymous
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.]
AP Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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rvc
  • rvc
r= 76m v= 100 km/h --> 100 X 5/18 m/s \(\mu\)= 0.38
anonymous
  • anonymous
@rvc it needs a max and a min /:
Michele_Laino
  • Michele_Laino
on our car are acting two forces, namely: |dw:1435475379190:dw|

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Michele_Laino
  • Michele_Laino
the condition for v_max, is then: \[\Large m\frac{{{v^2}}}{R} = \mu mg\]
Michele_Laino
  • Michele_Laino
which comes from the subsequent condition: \[\Large {F_{centrifugal}} = {F_{friction}}\]
Michele_Laino
  • Michele_Laino
solving the last expression for v, we get: \[\Large {v_{\max }} = \sqrt {\mu Rg} \]
Michele_Laino
  • Michele_Laino
actually our curve is, like below: |dw:1435475861915:dw|
rvc
  • rvc
v^2= r g tan \(\theta\)
Michele_Laino
  • Michele_Laino
we have this vector decomposition: |dw:1435476024271:dw|
Michele_Laino
  • Michele_Laino
so we can write: \[m\frac{{{v^2}}}{R}\cos \theta = \mu mg\]
Michele_Laino
  • Michele_Laino
\[\Large m\frac{{{v^2}}}{R}\cos \theta = \mu mg\]
rvc
  • rvc
i think it should be tan
Michele_Laino
  • Michele_Laino
using 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
  • Michele_Laino
why, tan?
Michele_Laino
  • Michele_Laino
|dw:1435476363488:dw|
Michele_Laino
  • Michele_Laino
|dw:1435476471088:dw|
Michele_Laino
  • Michele_Laino
you are right! @rvc
rvc
  • rvc
:)
Michele_Laino
  • Michele_Laino
the right friction force is: \[\Large \mu \left( {mg\cos \theta + m\frac{{{v^2}}}{R}\sin \theta } \right)\]
Michele_Laino
  • Michele_Laino
so 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
  • Michele_Laino
oops.. I forgot the component of the weight of the car along the inclined plane: \[\large m\frac{{{v^2}}}{R}\cos \theta = m\mu \left( {g\cos \theta + \frac{{{v^2}}}{R}\sin \theta } \right) + mg\sin \theta \]
rvc
  • rvc
look too big
Michele_Laino
  • Michele_Laino
if 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
  • Michele_Laino
or: \[\Large \tan \theta = \frac{1}{\mu }\]
Michele_Laino
  • Michele_Laino
the 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
  • Michele_Laino
which 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
  • Michele_Laino
solving the second equation for N, and substituting into the first equation, we get: \[\large m\frac{{{v^2}}}{R}\cos \theta - mg\sin \theta - \mu \left( {mg\cos \theta + m\frac{{{v^2}}}{R}\sin \theta } \right) = 0\]
rvc
  • rvc
cool
Michele_Laino
  • Michele_Laino
after simplification, we get: \[\Large {v^2} = gr\frac{{\mu + \tan \theta }}{{1 - \mu \tan \theta }}\] where r is the radius of our curve
rvc
  • rvc
yes!
Michele_Laino
  • Michele_Laino
we get the min value of v, if we set: \[\Large \tan \theta = 0\]
Michele_Laino
  • Michele_Laino
better is if we study this function, using the mathematical analysis: \[\Large f\left( \theta \right) = \frac{{\mu + \tan \theta }}{{1 - \mu \tan \theta }}\]
rvc
  • rvc
thats simpler :)
Michele_Laino
  • Michele_Laino
I 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
  • Michele_Laino
which is equivalent
Michele_Laino
  • Michele_Laino
I 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
  • rvc
well now let the user find out the equation
Michele_Laino
  • Michele_Laino
ok! 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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