+33653 Let load on front axle be L
Net upward force = 3L
Net downward force = W + 3000g (g+9.8m/s2)
W + 3000g = 3L ...(1)
W acts a horizontal distance (1/2)(4.32 + 2.16/2)m = 2.7m from the back of the cab; so take moments about the back wheel:
Clockwise: W*(3.6 - 2.7) + 3000g*(3.6 + 0.6) = 0.9W + 12600g
Anticlockwise: L*(1.8 + 0.6 + 3.6) = 6L
Equate these (assuming the truck is stable!)
0.9W + 12600g = 6L ...(2)
Now you have two simultaneous equations which you can solve for W and L.
I'll leave you to do this.
+33653 Let load on front axle be L
Net upward force = 3L
Net downward force = W + 3000g (g+9.8m/s2)
W + 3000g = 3L ...(1)
W acts a horizontal distance (1/2)(4.32 + 2.16/2)m = 2.7m from the back of the cab; so take moments about the back wheel:
Clockwise: W*(3.6 - 2.7) + 3000g*(3.6 + 0.6) = 0.9W + 12600g
Anticlockwise: L*(1.8 + 0.6 + 3.6) = 6L
Equate these (assuming the truck is stable!)
0.9W + 12600g = 6L ...(2)
Now you have two simultaneous equations which you can solve for W and L.
I'll leave you to do this.
