Answer:
The greatest possible acceleration of the car is [tex]a_G= 6.78 m/s^2[/tex]
Explanation:
[tex]N_A+N_B-Mg = 0[/tex]
[tex]-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0[/tex]
[tex]0.8N_B +0.8N_A = 975a_G[/tex]
[tex]N_A+N_B = 9564.75[/tex] -------------(1)
[tex]-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0[/tex]
[tex]-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0[/tex]
[tex]-2.26N_A -0.06N_B= 0[/tex] ----------------(2)
Solving the equation (1) and(2)
[tex]N_A + N_B = 9564.75[/tex]
[tex]-2.26N_A-0.06N_B=0[/tex]
[tex]N_A = -260.85N[/tex]
[tex]N_B = 9825.60N[/tex]
[tex]\mu_s N_B + \mu_s N_A = 975a_G[/tex]
[tex]0.8(9825.60)+0.8(-260.85) = 975a_G[/tex][tex]a_G=\frac{7651.8}{975}[/tex][tex]a_G_1=7.4848m/s^2[/tex]
Next lets assume that the front wheels contact with the ground N_A = 0
[tex]F_B = Ma_G[/tex]
[tex]N_B = M_g[/tex]
[tex]N_B - M_g = 0[/tex]
[tex]N_B(b-a) –F_Bh = 0[/tex]
[tex]F_B = 975a_G[/tex]
[tex]N_B-975(9.8) = 0[/tex]
[tex]N_B=9564.75N[/tex]
[tex]9564.75(2.20 -1.82) -F_B(0.55)=0[/tex]
[tex]\frac{3634.605}{0.55}=F_B[/tex]
[tex]F_B = 6608.3[/tex]
[tex]F_B = Ma_G[/tex]
[tex]6608.3 = 975a_G[/tex]
[tex]a_G = 6.7778 m/s^2[/tex]
[tex]a_G_2 = 6.78m/s^2[/tex]
Choosing the critical case
[tex]a_G = min(a_G_1 ,a_G_2)[/tex]
[tex]a_G = min(7.848, 6.78)[/tex]
[tex]a_G= 6.78 m/s^2[/tex]
