Question

part a
determine the magnitude and direction of the moment of the force at a about point p.
$m_p = 600\\ n\cdot m$?
$m_p = 1440\\ n\cdot m$?
$m_p = 600\\ n\cdot m$
$m_p = 1440\\ n\cdot m$

Explanation:

Step1: Resolve the force into components

The force of magnitude $F = 260$ N has components. The horizontal component $F_x=260\times\frac{12}{13}=240$ N and the vertical component $F_y = 260\times\frac{5}{13}=100$ N.

Step2: Calculate the moment about point $P$

The moment of a force about a point is given by $\vec{M}=\vec{r}\times\vec{F}$. The distance from point $P$ to point $A$ in the $x -$ direction is $d_x=2 + 4=6$ m and in the $y -$ direction is $d_y = 0$. The moment due to the horizontal component of the force about point $P$ is $M_{P_x}=F_y\times d_x$. Since $F_y = 100$ N and $d_x=6$ m, $M_{P_x}=100\times6 = 600$ N·m. The moment due to the vertical component about point $P$ is zero (because the line of action of the vertical - component passes through the $y -$ coordinate of point $P$). Using the right - hand rule, the direction of the moment is counter - clockwise.

Answer:

$M_P = 600$ N·m $\circlearrowleft$ (corresponding to the first option $M_P = 600$ N·m $\circlearrowleft$ where $\circlearrowleft$ represents counter - clockwise direction)