Coordinate System Diagram

Before Brickes are Raised

After Bricks are Raised

Use the support conventions presented in previous section to make a free-body diagram of the truck and bricks. Assume that the normal forces acting on the wheels, the center of mass of the truck, and the center of mass of the bricks all lie in the x-y plane.

Summing the forces in both the x and y direction, and the moments about point the center of gravity of the truck, point A, gives,

     ΣF_x = 0                                                       (1)
     ΣF_y = N₁ + N₂ - m₁g - m₂g = 0                     (2)
     ΣM_A = N₂ L₁ - N₁ L₁ - m₂g L₂ cosθ = 0         (3)

The first equation shows that there is no force exerted on the wheels by the ground in the horizontal direction. Using the third equation, solve for N₂,

     N₂ = N₁ + L₂/L₁ m₂g cosθ

and then substitute into the second equation,

     N₁ + N₁ + L₂/L₁ m₂g cosθ = g(m₁ + m₂)

     N₁ = g/2 (m₁ + m₂) - L₂/L₁ m₂g/2 cosθ

Solving

   N₁ = 9.81/2 (2,000 + 400)
               - (8/1.4) (400)(9.81)/2 cos30

   N₁ = 2,063 N

Substituting back into the second equation gives N₂ as,

     N₂ = g(m₁ + m₂) + N₁

     N₂ = 9.81 (2,000 + 400) - 2,063

     N₂ = 21,480 N

Notice that indeed

     N₁ + N₂ = 23,540 N

                  = g(m₁ + m₂)