The people mover shown weighs 20,000 lb. when empty. If the maximum allowable lo

Question

The people mover shown weighs 20,000 lb. when empty. If the maximum allowable load on any tire is 15,000 lb., what is the maximum number of people it should be rated to carry? Notes and Assumptions: Assume that: The average passenger weighs 150 lb. The passenger weight is uniformly distributed across the symmetrical passenger compartment such that their resultant load acts through the center of gravity, G. The empty weight of the people mover acts through its center of gravity as well. There are a total of six tires, 3 in front and 3 in back and the total applied load is equally distributed, front and back, between these two sets of tires. The tires only resist forces acting perpendicular to their tread. The people mover shown weighs 20,000 lb. when empty. If the maximum allowable load on any tire is 15,000 lb., what is the maximum number of people it should be rated to carry? Notes and Assumptions: Assume that: The average passenger weighs 150 lb. The passenger weight is uniformly distributed across the symmetrical passenger compartment such that their resultant load acts through the center of gravity, G. The empty weight of the people mover acts through its center of gravity as well. There are a total of six tires, 3 in front and 3 in back and the total applied load is equally distributed, front and back, between these two sets of tires. The tires only resist forces acting perpendicular to their tread.

Explanation / Answer

OR

taking moment about A

F2*4 = mg*5
=> F2 = (M+nm)g*5/4

F1 = F2 ( horizontal forces are cancelled for equilibrium)
=> F2 = (M+nm)g*5/3

(M+nm)g = F3 ( vertical forces are cancelled for equilibrium)

thus calculating:

(1)
15000*32.2 = [(20000 + n*150)*32.2*5/4]/2
=> n= 146.667

(3)

15000*32.2 < [20000+146.667*150]*32.2/2
<21000

thus not possible

so taking (3)

15000*32.2 = (150*n + 20000) * 32.2/2
=> n=66.667

thus number of people is 66 ANSWER

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