Question

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of F= (1.8v^2) N, where v is in m/s. Determine the deceleration the dragster can have upon releasing the parachute such that the wheels at B are on the verge of leaving the ground i.e., the normal reaction at B is zero. Neglect the mass of the wheels and assume the engine is disengaged so that the wheels are free to roll.G is 1.25m from A and B.

Answer 1

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a   ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²