Question

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

Answer 1

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

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

$\frac{3}{2}F_p = Mg$

$F_p = \frac{2}{3}Mg$

Answer 2

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