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wolverine [178]
3 years ago
15

"A bridge, constructed of 11 beams of equal length L and negligible mass, supports an object of mass M.

Physics
2 answers:
Vanyuwa [196]3 years ago
8 0

Answer:

force at P is 2/3 times of the weight

F_p = \frac{2}{3}Mg

Explanation:

here by force balance in vertical direction we can say

F_p + F_Q = Mg

now we can also apply torque balance for the beam as beam always remains horizontal

Torque due to Fp at the position of mass must be balanced by force due to Fq at the same position

(L)\times F_p = (L + L)\times F_Q

so from this equation we have

F_Q = \frac{1}{2}F_p

now from the first equation we have

F_p + \frac{1}{2}F_p = Mg

\frac{3}{2}F_p = Mg

F_p = \frac{2}{3}Mg

fiasKO [112]3 years ago
7 0
F_P  + F_Q = M g
F_P = M g - F_Q
Torque, or moment of force:
∑ M_P = 0
∑ M_P = M g L - F_Q · 3 L
0 = M g L - 3 F_Q L         / : L
0 = M g - 3 F_Q
3 F_Q = M g
F_Q = M g /3
Finally:
F_P = M g - M g/3
F_P = 4 M g / 3

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