# A concrete highway curve of radius 80.0 m is banked at a 11.0 degree angle.?

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What is the maximum speed with which a 1800 kg rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be 1.0.)

Please just give me a place to start instructor didn’t go over this and I am at a lose as to where to begin.

• Draw the free-body diagram of the car as seen from the back at the maximum velocity(it would resemble a block kept on an incline):

mg acts downwards

Normal reaction, R acts perpendicular to the highway

Friction, f acts down the plane

(We know from practical experience that when a body moves in a circle, it tends to move out of the circle more firmly as the velocity of the body is increased. So here the car is about to sip up the plane away from the center of the curve and hence friction acts down the plane.)

Resolve R and f into components.

At the maximum velocity, f is at its limiting value.

So f = uR (where u is the coefficient of friction)

Since acceleration in the vertical direction is 0

R cos11 – f sin11 = mg

or R cos11 – uR sin11 = mg …………(1)

In the horizontal direction, net acceleration is centripetal acceleration which is v^2 / r (r = 80 m)

So

R sin 11 + f cos 11 = mv^2 / r

or R sin 11 + uR cos 11 = mv^2 / r……….(2)

Divide (1) and (2)

v^2 / rg = (sin 11 + u cos 11) / (cos 11 – u sin11)

= (tan11 + u) / (1 – u tan11)

Substitute the values of r, g, u and tna11 and calculate v.

Hope this helps.

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• The financial company skill the line is gently grew to become to the area. this is like going sideways on a hill. you’re superb, the Fnet = Ff + Fn this implies the centripetal acceleration of the motor vehicle, the cost ought to equivalent the Ff + Fn to keep away from skidding. So, Ff = (coef. of friction) * Fn. because of the fact the coef. of friction = a million, the Ff + Fn = 2Ff So, Fnet = 2Ff Now, set Fnet = Fcentripetal acceleration initiate with Newton’s regulation, F=ma increasing, all of us comprehend acceleration a, on a curved direction = v^2/r so… Fcentripetal acceleration = mv^2/r Now substitute. mv^2/r = 2Ff (1900kg * v^2) / 60m = 2 * (a million.0 * 1900kg * 9.8m/s^2) (1900kg * v^2) / 60m = 2 * 18,620N v^2 = (one hundred twenty * 18,620N) / 1900kg v = 34.3 m/s

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