A bridge of length 75 m and mass 34000 kg is supported at each end (points A and B on the picture below). A truck of mass 18000 kg is located 27 m from the left support A.
Calculate the normal force on the bridge at point B (the right support). The acceleration of gravity is 9.8 m/s2 Answer in units of N. Calculate the normal force on the bridge at point A (the left support). Answer in units of N.

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karmenchong karmenchong · Answer 1 · 2024-01-25T09:57:58+01:00

Answer:

[tex]N_A=279496\ N[/tex]

Explanation:

Given:

length of bridge = 75 m

[tex]m_b = 34000\ kg[/tex]

[tex]m_t=18000\ kg[/tex]

distance A ↔ truck = 27 m

g = 9.8 m/s²

distance center of bridge ↔ B (d₁) = 75 ÷ 2

= 37.5 m

distance truck ↔ B (d₂) = 75 - 27

= 48 m

[tex]\Sigma F_y=0[/tex]

[tex]N_A+N_B-F_t-F_b=0[/tex]

[tex]N_A+N_B=F_t+F_b[/tex]

[tex]N_A+N_B=18000(9.8)+34000(9.8)[/tex]

[tex]N_A+N_B=509600\ N[/tex]

[tex]\tau_B=0[/tex]

[tex]N_B(0)+F_t\cdot d_1+F_b\cdot d_2-N_A(75)=0[/tex]

[tex]0+34000(9.8)(37.5)+18000(9.8)(48)-75N_A=0[/tex]

[tex]75N_A=20962200[/tex]

[tex]N_A=279496\ N[/tex]

[tex]N_A+N_B=509600[/tex]

[tex]279496+N_B=509600[/tex]

[tex]N_B=230104\ N[/tex]