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Use spacebar or mouse to advance slides. Problem 4.1 Applied Loads & Reactions Loading up that pickup truck!
A 1400-kg pickup truck is loaded with two crates, each having a mass of 350 kg. Determine the reactions at each of the two rear wheels and front wheels.
Consider just the pickup truck.
Its weight is concentrated at its center of gravity. 1.8 m1.2 m AB 0.75 m
This weight is distributed over the two axles, at R A and R B. 1.8 m1.2 m AB RARA RBRB
SUGGESTION: Since the loads are given in kilograms with even numbers, it might be best to calculate with kg and convert to Newtons at the end.
Analyze as though a simple beam. F = 0 = kg + RA RA + RBRB MB MB = 1400(1.2 m) - R A (3.0 m) RA RA = 560 kg x 9.81m/sec 2 and RB RB = 840 kg x 9.81m/sec m1.2 m AB 0.75 m RARA RBRB
The reactions at the wheels are: R A = 5.49 kN R B = 8.24 kN 1.8 m1.2 m AB 0.75 m RARA RBRB
Now let’s add some crates to the bed of the truck and determine the reactions at the wheels. 1.8 m1.2 m AB RARA RBRB
Each crate has a mass of 350 kg and is positioned as shown. 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD
Repeat the same calculations with the added loads as shown. 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD
F= 0 = RA RA + RBRB 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD 350 kg 1400 kg 350 kg
M B = 1400(1.2 m) - R A (3.0 m) + 350(2.05 m) + 350(3.75 m) 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD 350 kg 1400 kg 350 kg 2.05 m
M B = 1400(1.2 m) - R A (3.0 m) + 350(2.05 m) + 350(3.75 m) 1, ,312.5 = R A (3.0 m) 3,7100 kg-m = R A (3.0 m) 3,7100 kg-m = R A (3.0 m) R A = 1, kg x 9.81 m/sec 2 R A = kN
F= 0 = RA RA + RBRB R B = RARA RB RB = 2100 kg kg RB RB = kg x 9.81 m/sec 2 RB RB = 8.47 kN AB R A = kNR B = 8.47 kN 350 kg 1400 kg 350 kg
Final results are shown. Each rear wheel = 6.07 kN Each front wheel = 4.24 kN AB R A = kNR B = 8.47 kN 350 kg 1400 kg
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