a 1250 kg car is traveling around a banked curve at a speed of 23.5 m/s. the cur

Question

a 1250 kg car is traveling around a banked curve at a speed of 23.5 m/s. the curve is banked 24 degrees to the horizontal. the car does no depend on the force of friction to keep it on track. ignore air resistance. a) draw a free body diagram for the car, including all forces acting on it. clearly label and name each force. b) which force components, if any, are balanced? c) which force components, if any, are unbalanced? d) use your results from part c to derive an expression for the radius of the curve that depends on the speed of the car, the gravitational acceleration (g), and the angle that the curve is banked. e) what is the radius of the curve? please show clear and simple steps and formulas, thanks!!!

Explanation / Answer

The normal force (n) of the car is shifted from the vertical (y axis) to a degree equal to the banking and mustnow be analyzed in terms of its x and y components. Draw a right angle triangle from above the banking to represent the x and y vectors of the normal force.

a/b/c) In order for the car to be in equilibrium the horizontal and vertical componenet must be in equilibrium. The x component of the normal force n(x) must be equal to the radial force F(r),and the y component of the normal force n(y) must be equal to the force of gravity.

F(r) = n(x)
ma(c) = n sin(?)

F(g) = n(y)
mg = n cos(?)

Since we knowthe angle of the banking, we have enough information to solve for (n).
n = mg / cos(?)

We can now solve for the centripetal acceleration by substituting the value of (n).

ma(c) = n sin(?)
= [mg / cos(?)] sin(?)
= mg sin(?) / cos(?)
a(c) = g tan(?)

d/e.) Substitute the formula for centripetal acceleration and solve for the desired variable, which in this case, is (r) radius of the turn.

V^2 / r = g tan(?)
r = v^2 / g tan(?)
= (23.5 m/s)^2 / (9.8 m/s^2)tan(24.0?)
= 127 m

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