Question

suppose that $f_1 = 240\\ n$ and $f_2 = 450\\ n$. part h in the case (h) (figure 8), determine the moment of the force about point o. express your answer to three significant figures and include the appropriate units. enter positive value if the moment is counterclockwise and negative value if the moment is clockwise. part i in the case (i) (figure 9), determine the moment of the force about point o. express your answer to three significant figures and include the appropriate units. enter positive value if the moment is counterclockwise and negative value if the moment is clockwise.

Explanation:

Step1: Resolve force $F_1$ into components

The vertical component of $F_1$, $F_{1y}=F_1\times\frac{3}{5}$, and the horizontal component $F_{1x}=F_1\times\frac{4}{5}$. Given $F_1 = 240$ N, so $F_{1y}=240\times\frac{3}{5}=144$ N and $F_{1x}=240\times\frac{4}{5}=192$ N.

Step2: Calculate moment due to components of $F_1$ about point $O$

The moment due to $F_{1y}$ about point $O$ is $M_{1y}=F_{1y}\times(1 + 1)=144\times2 = 288$ N·m (counter - clockwise). The moment due to $F_{1x}$ about point $O$ is $M_{1x}=F_{1x}\times3=192\times3 = 576$ N·m (counter - clockwise).

Step3: Calculate moment due to $F_2$ about point $O$

$F_2 = 450$ N acts at a perpendicular distance of $1$ m from point $O$. So the moment due to $F_2$ about point $O$ is $M_2=F_2\times1=450\times1 = 450$ N·m (clockwise).

Step4: Calculate the net moment $M_O$

$M_O = M_{1y}+M_{1x}-M_2$. Substituting the values, we get $M_O=288 + 576-450=414$ N·m.

Answer:

$414$ N·m