Answer:
[tex]437.1053\,{\rm{lb}}\\\end{array}[/tex]
Explanation:
Assuming the figure attached with its dimensions
The normal force under the front and rear wheels are equal.
[tex]{R_A} = {R_B} = R[/tex]
Apply force equilibrium equation.
[tex]\begin{array}{l}\\\Sigma F = 0\\\\{R_A} + {R_B} - W - {W_L} = 0\\\\\\\end{array}[/tex]
Substitute 3020 lb for W and R for [tex]R_A[/tex] and [tex]R_B[/tex]
[tex]\begin{array}{l}\\R + R - 3020 - {W_L} = 0\\\\2R = 3020 + {W_L}\\\\R = \frac{1}{2}\left( {3020 + {W_L}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....(1)\\\end{array}[/tex]
Take moment about point [tex]W_L[/tex]
[tex]\begin{array}{l}\\\Sigma {M_w} = 0\\\\ - {R_A} \times \left( {45 + 67 + 20} \right) + W \times \left( {67 + 20} \right) - {R_B}\left( {20} \right) = 0\\\\ - 132{R_A} + 87W - 20{R_B} = 0\\\end{array}[/tex]
[tex]\begin{array}{l}\\ - 132 \times R + 87 \times 3020 - 20 \times R = 0\\\\ - 152R + 262740 = 0\\\end{array}[/tex]
Substitute [tex]\frac{1}{2}\left( {3020 + {W_L}} \right) [/tex] for R from equation (1).
[tex]\begin{array}{l}\\ - 152 \times \left( {\frac{1}{2}\left( {3020 + {W_L}} \right)} \right) + 262740 = 0\\\\\left( {3020 + {W_L}} \right) = 3457.105\\\\{W_L} = 437.1053\,{\rm{lb}}\\\end{array}[/tex]
