1. Problem 4.1 Applied Loads & Reactions
Loading up that pickup truck!
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2. 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.

3. Consider just the pickup truck.

4. Its weight is concentrated at its center of gravity.B
1.8 m
1.2 m
0.75 m

5. This weight is distributed over the two axles, at RA and RB.
1.8 m
1.2 m
0.75 m RB

6. 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.

7. Analyze as though a simple beam
Analyze as though a simple beam. F = 0 = kg + RA + RB MB = 1400(1.2 m) - RA(3.0 m)  RA = 560 kg x 9.81m/sec2 and RB = 840 kg x 9.81m/sec2 A B RA
1.8 m
1.2 m
0.75 m RB

8. The reactions at the wheels are: RA = 5.49 kN RB = 8.24 kN
1.8 m
1.2 m
0.75 m RB

9. Now let’s add some crates to the bed of the truck and determine the reactions at the wheels.RB

10. Each crate has a mass of 350 kg and is positioned as shown.B RA
1.8 m
1.2 m
0.75 m RB

11. Repeat the same calculations with the added loads as shown.B RA
1.8 m
1.2 m
0.75 m RB

12. F= 0 = -350 -350 -1400 + RA + RB RA RB 1.7 m 2.8 m C D 350 kg 350 kg

13. MB = 1400(1.2 m) - RA(3.0 m) + 350(2.05 m) + 350(3.75 m)C D
350 kg
350 kg
1400 kg A B RA
1.8 m
1.2 m
0.75 m RB

14. MB = 1400(1.2 m) - RA(3.0 m) + 350(2.05 m) + 350(3.75 m)
1, , = RA(3.0 m) 3,7100 kg-m = RA(3.0 m) 3,7100 kg-m = RA (3.0 m)
RA = 1, kg x 9.81 m/sec2
RA = kN

15. F= 0 = -350 -350 -1400 + RA + RB. RB = 350 + 350 + 1400 - RA
F= 0 = RA + RB RB = RA RB = 2100 kg kg RB = kg x 9.81 m/sec2 RB = 8.47 kN
350 kg
350 kg
1400 kg A B
RA = kN
RB = kN

16. Final results are shown. Each rear wheel = 6
Final results are shown. Each rear wheel = 6.07 kN Each front wheel = 4.24 kN
350 kg
350 kg
1400 kg A B
RA = kN
RB = kN