The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft diameter circle painted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If the maximum speed is 35 mi/hr for a 3000-lb car, compute the magnitude F of the total friction force exerted by the pavement on the car tires.

Question

Answered

Answer :

NUHULAWAL20GO NUHULAWAL20GO · Answer 1

Answer:

the magnitude F of the total friction force is 2456.7 lb

Explanation:

Given the data in the question;

maximum speed = 35 mi/hr = ( 35×5280 / 60×60) = 51.3333 ft/s

diameter = 200ft

radius = 200/2 = 100 ft

First we calculate the normal component of the acceleration;

[tex]a_{n}[/tex] = v² / p

where v is the velocity of the car( 51.3333 ft/s)

p is the radius of the curvature( 100 ft)

so we substitute

[tex]a_{n}[/tex] = (51.3333 ft/s)² / 100ft

[tex]a_{n}[/tex] = (2635.1076 ft²/s²) / 100ft

[tex]a_{n}[/tex] = 26.35 ft/s²

we convert Feet Per Second Squared (ft/s²) to Standard Gravity (g)

1 ft/s² = 0.0310809502 g

[tex]a_{n}[/tex] = 26.35 ft/s² × 0.0310809502 g

[tex]a_{n}[/tex] = 0.8189g

Now consider the dynamic equilibrium of forces in the Normal Direction;

∑[tex]F_{n}[/tex] = m[tex]a_{n}[/tex]

F = m[tex]a_{n}[/tex]

we know that mass of the car is 3000-lb = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb)

so

we substitute

F = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb) × 0.8189g

F = 2456.7 lb

Therefore; the magnitude F of the total friction force is 2456.7 lb