1. **State the problem:** We have a truck with mass 4300 kg and a pallet of shingles with mass 1600 kg on a boom. We need to find the reaction forces at the rear wheels B and front wheels C. 2. **Given data:** - Mass of truck $m_t = 4300$ kg - Mass of pallet $m_p = 1600$ kg - Boom length $L = 6$ m - Boom angle $\theta = 15^\circ$ - Truck length components: $4.3$ m + $0.5$ m = $4.8$ m total - Gap near front wheels: $0.4$ m 3. **Assumptions and simplifications:** - Gravity $g = 9.8$ m/s$^2$ - The truck weight acts at its center of gravity, assumed at midpoint of truck length: $4.8/2 = 2.4$ m from rear wheels B - The pallet weight acts at the end of the boom, which is $6$ m from the pivot point - The pivot point is located at the rear wheels B 4. **Calculate weights:** $$ W_t = m_t \times g = 4300 \times 9.8 = 42140 \text{ N} $$ $$ W_p = m_p \times g = 1600 \times 9.8 = 15680 \text{ N} $$ 5. **Calculate horizontal distance of pallet load from rear wheels B:** The boom is inclined at $15^\circ$, so horizontal projection: $$ x_p = 6 \times \cos 15^\circ = 6 \times 0.9659 = 5.7954 \text{ m} $$ 6. **Sum of vertical forces:** Let $R_B$ be reaction at rear wheels B, $R_C$ at front wheels C. $$ R_B + R_C = W_t + W_p = 42140 + 15680 = 57820 \text{ N} $$ 7. **Sum of moments about rear wheels B (taking counterclockwise as positive):** Moments due to weights and reactions: - Truck weight acts at 2.4 m from B: moment $= W_t \times 2.4$ - Pallet weight acts at $x_p + 0.4 + 0.5 = 5.7954 + 0.4 + 0.5 = 6.6954$ m from B (adding gap and front wheel offset) - Reaction at front wheels C acts at $4.8 + 0.4 = 5.2$ m from B Moment equation: $$ W_t \times 2.4 + W_p \times 6.6954 - R_C \times 5.2 = 0 $$ 8. **Solve for $R_C$:** $$ R_C = \frac{W_t \times 2.4 + W_p \times 6.6954}{5.2} = \frac{42140 \times 2.4 + 15680 \times 6.6954}{5.2} $$ Calculate numerator: $$ 42140 \times 2.4 = 101136 $$ $$ 15680 \times 6.6954 = 104993.47 $$ Sum: $$ 101136 + 104993.47 = 206129.47 $$ Divide: $$ R_C = \frac{206129.47}{5.2} = 39640.28 \text{ N} $$ 9. **Calculate $R_B$ using vertical force balance:** $$ R_B = 57820 - 39640.28 = 18179.72 \text{ N} $$ **Final answers:** - Reaction at rear wheels B: $\boxed{18180 \text{ N}}$ - Reaction at front wheels C: $\boxed{39640 \text{ N}}$