Question

The indicated location of the center of gravity of the 3020-lb pickup truck is for the unladen condition. If a load whose center of gravity is x = 20 in. behind the rear axle is added to the truck, determine the load weight WL for which the normal forces under the front and rear wheels are equal.

Answer 1

`437.1053\,{\rm{lb}}\\\end{array}`

Step-by-step explanation:

Assuming the figure attached with its dimensions

The normal force under the front and rear wheels are equal.

`{R_A} = {R_B} = R`

Apply force equilibrium equation.

`\begin{array}{l}\\\Sigma F = 0\\\\{R_A} + {R_B} - W - {W_L} = 0\\\\\\\end{array}`

Substitute 3020 lb for W and R for
`R_A` and
`R_B`

`\begin{array}{l}\\R + R - 3020 - {W_L} = 0\\\\2R = 3020 + {W_L}\\\\R = (1)/(2)\left( {3020 + {W_L}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....(1)\\\end{array}`

Take moment about point
`W_L`

`\begin{array}{l}\\\Sigma {M_w} = 0\\\\ - {R_A} * \left( {45 + 67 + 20} \right) + W * \left( {67 + 20} \right) - {R_B}\left( {20} \right) = 0\\\\ - 132{R_A} + 87W - 20{R_B} = 0\\\end{array}`

`\begin{array}{l}\\ - 132 * R + 87 * 3020 - 20 * R = 0\\\\ - 152R + 262740 = 0\\\end{array}`

Substitute
`(1)/(2)\left( {3020 + {W_L}} \right)` for R from equation (1).

`\begin{array}{l}\\ - 152 * \left( {(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}`