In the figure, the driver of a car on a horizontal road makes an emergency stop

Question

In the figure, the driver of a car on a horizontal road makes an emergency stop by applying the brakes so that all four wheels lock and skid along the road. The coefficient of kinetic friction between tires and road is 0.43. The separation between the front and rear axles is L = 4.6 m, and the center of mass of the car is located at distance d = 1.6 m behind the front axle and distance h = 0.77 m above the road. The car weighs 10 kN. Find the magnitude of (a) the braking acceleration of the car, (b) the normal force on each rear wheel, (c) the normal force on each front wheel, (d) the braking force on each rear wheel, and (e) the braking force on each front wheel.

Explanation / Answer

a )

Given weight of the car W = 10 kN

= 10000 N

but we have W = m g = 10000

then finding the mass of car is m = W / g

= 10000 / 9.8 = 1020.40 Kg

we have force F = m a = - m g

and = 0.5

now finding the accelaration of the car is a

a = - g

= 0.5 X 9.8

= - 4.9 m / s2

b )

frictional force on the car f = m g

= 0.5 X 10000 N

= 5000 N

normal force on car on rear wheel is Fnr

normal force on car on front wheel is Fnf

and with using given data h = 0.77 m

L = 4.6 m , d = 1.9 m

total force is Fnr + Fnf = m g ..........1

Fnr + Fnf = 10000...............2

and f X h + Fnr X ( L - d ) - Fnf X d = 0............3

5000 X 0.77 + Fnr ( 4.6 - 1.6 ) - Fnf X 1.6 = 0 .........4

to get the forces solving the equations 2 and 4

3850 + Fnr X 3 - (10000 - Fnr ) X 1.6 = 0

3850 + Fnr X 3 - (10000 X 1.6) + 1.6 Fnr = 0

4.6 Fnr = 12150

Fnr = 2641.30 N

Fnf = 10000 - 2641.30

Fnf = 7358.70 N

the normal force on each rear wheel is Fnr

= Fnr / 2 ( beacuse rear wheels are two)

= 2641.30 / 2

= 1320.65 N

c )

Fnf = 7358.70 N

and normal force on each front wheel is Fnf

= 7358.70 / 2 ( because front wheels also two )

= 3679.35 N

d )

breaking force on each rear wheel is frear = Fnr

= 0.5 X 1320.65

= 660.325 N

e )

breaking force on each front wheel is ffront = Fnf

= 0.5 X 3679.35

= 1839.675 N

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