A 1196-kg car and a 2010-kg pickup truck approach a curve on the expressway that

Question

A 1196-kg car and a 2010-kg pickup truck approach a curve on the expressway that has a radius of 253 m .

At what angle should the highway engineer bank this curve so that vehicles traveling at 60.0 mi/h can safely round it regardless of the condition of their tires?

Should the heavy truck go slower than the lighter car?

As the car and truck round the curve at 60.0 mi/h , find the normal force on the car to the highway surface

As the car and truck round the curve at 60.0 mi/h , find the normal force on the truck to the highway surface

Explanation / Answer

There are two forces acting on the vehicle. The downward force due to gravity equal to the weight of the vehicle

Fy = mg

and the horizontal force (centrifugal) force due to the speed and curvature of the road

Fx = m(v^2)/R

The road should be sloped so that the vector sume of these forces is perpendicular to the road surface

Tan[ang] = Fx/Fy = (v^2)/(R*g)

Note the vehicle mass does not appear in the above equation, so the result is independent of vehicle mass

So, tan (ang)=(60*0.447)^2/253*9.8=>

ang=tan^-1(0.29)
NOTICE it does not matter what the mass or weight of the vehicle is. For a given radius, it is only dependent on the speed.

b) Since the forces are perpendicular to the surface. and the two forces are perpendicular to each other. the normal force is

Fn = Sqrt[Fx^2 + Fy^2]

Note, the masses do not cancel here, so the normal force will be proportional to the masses.

For car, Fy=mg=1196*9.8=11720.8

and Fx=m(v^2)/R=1196*(60*0.446)^2/253=3385.18N

So normal force= (11720.8^2+3385.18^2)^0.5=12199.86 N

Similarly for truck,

Normal force=((2010*9.8)^2+(2010*(60*0.446)^2/253)^2)^0.5=20503.11N

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