\"Physics: Principles with Applications 6th Edition\" Chapter 5, Problem 22P sta
ID: 2061167 • Letter: #
Question
"Physics: Principles with Applications 6th Edition" Chapter 5, Problem 22P states:A 1200 kg car rounds a corner of radius 67 m, banked at an angle of 12 degrees. If the car is traveling at 95 km/h (26.4 m/s), will friction force be required? If so, how much and in what direction?
In step 2 of the solution, it says: "But applying (Sigma)Fy=0"
Can someone please explain how you can have no y value. If we're only concerned with a force in the horizontal (centripedal) direction, which triangle am I trying to solve? There are two in the drawing.
Explanation / Answer
We are actually not worried about the motion in the x or y direction directly. The car will not go in either of thise directions, The car will either slide up the plane or down the plane is the total forces are not equalized. Don't worry about anything solved in the y direction.
Centripatal force will pull directly into the center of the circle, and it will have a reaction force from Newtons Third Law that acts outward, call it a centrifugal force if you like.
That centrufugal force has a component up the plane
Fc = mv2/r
The component up the plane is (mv2/r)(cos12) = [(1200)(26.4)2/67)](cos 12) = 12210 N up the plane.
The car will have a component of its weight acting down the plane. That is mgcos
The weight component is (1200)(9.8)(cos12) = 11503 N down the plane
Since the two forces do not equalize, you do need a third force or the car would move along the plane. Since the force up is greater than the force down, the car would slide up the plane. So we need friction to act down the plane. The amount of friction needed is
12210 - 11503 = 707 N
Therefore we need 707 N worth of frictional force to act down the plane.
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