Question

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

Throughout this problem, use g for the magnitude of the acceleration due to gravity.
Find F_P, the vertical force that pier P exerts on the left end of the bridge.

"

Answer 1

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

Answer 2

force at P is 2/3 times of the weight

`F_p = (2)/(3)Mg`

Step-by-step 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)* F_p = (L + L)* F_Q`

so from this equation we have

`F_Q = (1)/(2)F_p`

now from the first equation we have

`F_p + (1)/(2)F_p = Mg`

`(3)/(2)F_p = Mg`

`F_p = (2)/(3)Mg`