Question

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

Answer 1

The greatest possible acceleration of the car is
`a_G= 6.78 m/s^2`

Step-by-step explanation:

`N_A+N_B-Mg = 0`

`-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0`

`0.8N_B +0.8N_A = 975a_G`

`N_A+N_B = 9564.75` -------------(1)

`-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0`

`-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0`

`-2.26N_A -0.06N_B= 0` ----------------(2)

Solving the equation (1) and(2)

`N_A + N_B = 9564.75`

`-2.26N_A-0.06N_B=0`

`N_A = -260.85N`

`N_B = 9825.60N`

`\mu_s N_B + \mu_s N_A = 975a_G`

`0.8(9825.60)+0.8(-260.85) = 975a_G`
`a_G=(7651.8)/(975)`
`a_G_1=7.4848m/s^2`

Next lets assume that the front wheels contact with the ground N_A = 0

`F_B = Ma_G`

`N_B = M_g`

`N_B - M_g = 0`

`N_B(b-a) –F_Bh = 0`

`F_B = 975a_G`

`N_B-975(9.8) = 0`

`N_B=9564.75N`

`9564.75(2.20 -1.82) -F_B(0.55)=0`

`(3634.605)/(0.55)=F_B`

`F_B = 6608.3`

`F_B = Ma_G`

`6608.3 = 975a_G`

`a_G = 6.7778 m/s^2`

`a_G_2 = 6.78m/s^2`

Choosing the critical case

`a_G = min(a_G_1 ,a_G_2)`

`a_G = min(7.848, 6.78)`

`a_G= 6.78 m/s^2`