The sports car has a weight of 4500 lb and center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft>s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are ms = 0.5 and mk = 0.3, respectively. Neglect the mass of the wheels

It takes 2.72 seconds for it to reach a speed of 10 ft/s

The normal reactions at each of the four wheels on the road are

NA = 1392.86 lb and NB = 857.14 lb

Explanation:

Given data;

Weight of the car = 4500 lb

Static coefficient of friction = 0.5

Kinetic coefficient of friction = 0.3

free body diagram of car is shown in the attached file.

Note:

normal reaction at wheel A is NA

normal reaction at wheel B is NB

Acceleration = ag

Form Force F = mass m * acceleration a

m = F/a

m = 4500/32.2 = 139.75 lb

Taking moment about point A

£MA = 0

-(2NB*6) + (4500*2) + (139.75*ag *2.5) = 0

-12NB +9000 + 349.38*ag = 0

349.38*ag - 12NB = -9000. ----eqn 1

Taking £Fx = 0

(0.3*2NB) - m*ag = 0

0.6NB - 139.75*ag = 0

0.6NB = 139.75*ag

ag = 0.6NB/139.75 ------eqn 2

Substitute eqn 2 into eqn 1

349.38*(0.6NB/139.75) - 12NB = -9000

(209.63NB/139.75) - 12NB = -9000

1.5NB - 12NB = -9000

-10.5NB = -9000

NB = 9000/10.5 = 857.14 lb

Also, taking £Fy = 0

-4500 + 2NB + 2NA = 0 ---- eqn 3

Substitute NB = 857.14 b into eqn 3

-4500 + 2(857.14) + 2NA = 0

-4500 + 1714.29 + 2NA = 0

-2785.71 + 2NA = 0

2NA = 2785.71

NA = 2785.71/2 = 1392.86 lb

To how long it takes for it to reach a speed of 10 ft/s

Find ag from eqn 1

ag = 0.6NB/139.75

ag = (0.6 * 857.14)/139.75 = 3.68 ft/s²

Using V = U + ag*t

Where t is time in seconds

U is initial velocity = O

V is finally velocity = 10 ft/s

10 = 3.68*t

t = 10/3.68 = 2.72 sec.

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: