Question

part b
as shown, a pipe is anchored to a wall at point a. during the pipes installation, several forces are applied to the pipe at different locations. if $f_1 = 15.9$ lb, $f_2 = 13.9$ lb, $f_3 = 13.2$ lb, $f_4 = 18.0$ lb, $d_1 = 0.300$ ft, $d_2 = 0.850$ ft, and $d_3 = 0.900$ ft, what is $m_{r_a}$, the net moment about point a due to these forces? assume that moments acting counterclockwise about point a are positive whereas moments acting clockwise are negative.
express your answer numerically in pound - feet to three significant figures.
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Explanation:

Step1: Calculate moment of $F_1$

The moment of a force $F$ about a point is $M = F\times d$. For $F_1$, the moment $M_1=-F_1\times d_2$ (clock - wise, so negative). Substituting $F_1 = 15.9$ lb and $d_2=0.850$ ft, we get $M_1=-15.9\times0.850=-13.515$ lb - ft.

Step2: Calculate moment of $F_2$

$F_2$ creates a counter - clockwise moment. So $M_2 = F_2\times d_1$. Substituting $F_2 = 13.9$ lb and $d_1 = 0.300$ ft, we get $M_2=13.9\times0.300 = 4.17$ lb - ft.

Step3: Calculate moment of $F_3$

$F_3$ creates a counter - clockwise moment. So $M_3=F_3\times d_3$. Substituting $F_3 = 13.2$ lb and $d_3 = 0.900$ ft, we get $M_3=13.2\times0.900=11.88$ lb - ft.

Step4: Calculate moment of $F_4$

$F_4$ creates a counter - clockwise moment. So $M_4=F_4\times d_3$. Substituting $F_4 = 18.0$ lb and $d_3 = 0.900$ ft, we get $M_4=18.0\times0.900 = 16.2$ lb - ft.

Step5: Calculate the net moment $M_{R_A}$

$M_{R_A}=M_1 + M_2+M_3+M_4$. Substituting the values of $M_1,M_2,M_3,M_4$ we found above: $M_{R_A}=-13.515 + 4.17+11.88+16.2$.
$M_{R_A}=(-13.515)+(4.17 + 11.88+16.2)=(-13.515)+32.25=18.735\approx18.7$ lb - ft.

Answer:

$18.7$ lb - ft