$R = W_r + W_f$

$3000 = 1800 + W_f$

$W_f = 1200 \, \text{lb}$

$Rx = 9W_f$

$3000x = 9(1200)$

$x = 3.6 \, \text{ft}$

$9 - x = 5.4 \, \text{ft}$

When the midspan is midway between W_{r} and R, the front wheel W_{f} will be outside the span (see figure). In this case, only the rear wheel W_{r} = 1800 lb is the load. The maximum moment for this condition is when the load is at the midspan.

$R_1 = R_2 = \frac{1}{2}(1800)$

$R_1 = 900 \, \text{lb}$

**Maximum moment under W**_{r}

$M_\text{To the left of rear wheel} = 7R_1$

$M_\text{To the left of rear wheel} = 7(900)$

$M_\text{To the left of rear wheel} = 6300 \, \text{lb}\cdot\text{ft}$

**Maximum moment under W**_{f}

$\Sigma M_{R1} = 0$

$14R_2 = 4.3R$

$14R_2 = 4.3(3000)$

$R_2 = 921.43 \, \text{lb}$

$M_\text{To the right of front wheel} = 4.3R_2$

$M_\text{To the right of front wheel} = 4.3(921.43)$

$M_\text{To the right of front wheel} = 3962.1 \, \text{lb}\cdot\text{ft}$

Thus,

$M_{max} = M_\text{To the left of rear wheel}$

$M_{max} = 6300 \, \text{lb}\cdot\text{ft} \,\,$ *answer*

**The maximum shear will occur when the rear wheel (wheel of greater load) is directly over the support.**

$\Sigma M_{R2} = 0$

$14R_1 = 10.4R$

$14R_1 = 10.4(3000)$

$R_1 = 2228.57 \, \text{lb}$

Thus,

$V_{max} = 2228.57 \, \text{lb} \,\,$ *answer*