The 1400 kg rear wheel drive car is hinged to a trailer having a total mass of 550 kg via a spherical joint at A. The car driver suddenly applies the breaks causing the car to skid on the road for which the coefficient of static friction with the rear tires is 0.59. Determine the car deceleration, and the horizontal and vertical components of reaction at A.

hello your question is incomplete attached below is the complete question

i)  = - 15.407 m/s^2  ( deceleration )

ii) Rax = 124.08 KN ( Horizontal reaction component )

Ray = 28.72 KN  ( vertical reaction component )

Explanation:

Given data :

Total mass of trailer = 550 kg

mass of rear wheel drive car = 1400 kg

coefficient of static friction = 0.59

i) determine the deceleration of the car

first we have to take moments about point A

Ma  = (-f * 0.4) + ( mg * 2 ) + ma(0.75 - 0.4 ) = 0 ----------- ( 1 )

f = u N  and N = mg  ( input this into equation 1 above )

equation 1 becomes

-u mg * 0.4 = -mg (2) - ma( 0.35 )

- 0.59 ( 1400 * 9.81 ) * 0.4 = - 550 ( 9.81 * 2) - 1400 (a) * 0.35

hence (a ) = ( - 10791 + 3241.334 ) / ( 1400 * 0.35 )

                = - 15.407 m/s^2  ( deceleration )

ii) Horizontal and vertical component of reaction A

we take moments about G and G1

MG1 =  Rax (1.25 - 0.4 ) - Ray (2) + mg (3.5) = 0

 Rax( 0.85 ) - 2Ray = 48020 -------- (2)

taking moments about G

MG = -Mgx2 + Ray*3.5 - Rax(0.35) - mg(5.5) = 0

   Ray * 3.5 - Rax * 0.35 = 57115.25 ---------- (3)

solve equation 2 and 3 simultaneously

Rax = 124.08 KN ( Horizontal components )

Ray = 28.72 KN  ( vertical components )

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