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If a curve with a radius of 85 is perfectly banked for a car traveling 71 , what must be the coefficient of static friction for...Asked by Emily Grossman
If a curve with a radius of 88m is perfectly banked for a car traveling 75km/h, what must be the coefficient of static friction for a car not skid when traveling at 95 km/h?
Answers
Answered by
bobpursley
change velocities to m/s
Then, sketch the bank and force vectors parallel to the bank.
1) mg*sinTheta=mv^2/r * cosTheta
where v=75km/hr in m/s
solve for tan theta (sin/cos).
verify that. Now, you increase speed, same angle.
mg*sinTheta+mu*mg*cosTheta=mv^2/r * cosTheta
so find mu. How is the easy way? divid e by cos theta , you know tan theta.
v= new speed (95 in m/s), solve for mu.
Then, sketch the bank and force vectors parallel to the bank.
1) mg*sinTheta=mv^2/r * cosTheta
where v=75km/hr in m/s
solve for tan theta (sin/cos).
verify that. Now, you increase speed, same angle.
mg*sinTheta+mu*mg*cosTheta=mv^2/r * cosTheta
so find mu. How is the easy way? divid e by cos theta , you know tan theta.
v= new speed (95 in m/s), solve for mu.
Answered by
Emily Grossman
Can you show me step by step with the numbers in there? I am confused..
Answered by
LB
-0.327
Answered by
Greg
You forgot the normal force due to centripetal force.
mg*sinTheta+mu*mg*cosTheta+mu*v^2/r*sinTheta=mv^2/r * cosTheta
m cancels out so the equilibrium is stated in acceleration.
mu= ((v^2/r)*cosTheta-g*sinTheta)/(g*cosTheta+(v^2/r)*sinTheta)
mg*sinTheta+mu*mg*cosTheta+mu*v^2/r*sinTheta=mv^2/r * cosTheta
m cancels out so the equilibrium is stated in acceleration.
mu= ((v^2/r)*cosTheta-g*sinTheta)/(g*cosTheta+(v^2/r)*sinTheta)
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