Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_31

⚡ Mecademy AIENG정역학 · ch3  Problem Statement The indicated location of the center of gravity of the 3600-lb pickup truck is for the unladen condition. If a load whose center of gravity is x = 16 in. behind the rear axle is added to the truck, determine the load weight for which the normal forces under the front and rear wheels are equal. Problem 3/31 (a) Determination of load weight for equal normal forces ● Calculation Process 1. Present Final Formula: To find the weight , we use the equilibrium equations for a rigid body. Let be the total normal force on the front wheels and be the total normal force on the rear wheels. For the wheels to have equal normal forces, we set . Sum of vertical forces: Sum of moments about the front axle : Where: (Weight of unladen truck) (Distance from front axle to truck mass center ) (Wheelbase) (Distance of load behind rear axle) 2. Substitute Values: Substitute into the moment equation: 3. Partial Operations: Simplify the equation: W L W L W L N A N B N = A N = B N F = ∑ y 0⟹2N=W+W L A M = ∑ A 0⟹N ⋅ B L−W⋅d − 1 W ⋅ L (L+x)=0 W=3600 lb d = 1 45 in.G L=45+67=112 in. x=16 in. N= 2 W+W L ⋅ ( 2 3600+W L )112−3600⋅45−W ⋅ L (112+16)=0 56(3600+W )− L 162000−128W = L 0 Combine like terms: 4. Final Calculation: ● Final Conclusion: The weight of the added load required to make the normal forces under the front and rear wheels equal is . ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 3/31 201600+56W − L 162000−128W = L 0 (201600−162000)=(1