Question

A gardener uses a 60-N wheelbarrow to transport two bags of fertilizer weighing W = 252-N. Determine the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm. (Round the final answer to three decimal places.)

Answer 1

Explanation:

Let us assume that the maximum allowable horizontal distance be represented by "d".  

Therefore, torque equation about A will be as follows.

   $60 \times 0.15 + 252 \times 0.15 \times 2 + 252 \times d = 2 \times 75 \times (0.7 + 0.15 + 0.15)$

      d = $\frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}$

       d = 0.409 m

Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.