(07) If the algebraic measure of the moment of force \( \vec{F} \) about each of the point \( \mathrm{O}(0,0), \mathrm{D}(1,0), \mathrm{E}(0,2) \) equals \( 27,18,40 \frac{1}{2} \) moment unit, find: \( \vec{F} \)

To solve for the force vector \( \vec{F} \) based on the given moments about the points, we start by establishing the relationship between the moment of a force and the position vector. The moment \( \vec{M} \) about a point \( \vec{r} \) due to a force \( \vec{F} \) applied at \( \vec{r}_F \) is given by: \[ \vec{M} = \vec{r} \times \vec{F} \] 1. **For point O(0,0)**: The moment is given as 27 units, which can be expressed as: \[ M_O = x_F F_y - y_F F_x = 27 \] 2. **For point D(1,0)**: The moment here is 18 units, leading to the equation: \[ M_D = (x_F - 1) F_y - y_F F_x = 18 \] 3. **For point E(0,2)**: The moment is given as \( 40 \frac{1}{2} = 40.5 \) units: \[ M_E = x_F F_y - (y_F - 2) F_x = 40.5 \] Now we can have a system of equations based on the above moments. Let's denote \( F_x \) and \( F_y \) as the components of the force \( \vec{F} \). From the moments, we get the following equations: 1. \( -y_F F_x + x_F F_y = 27 \) (1) 2. \( -y_F F_x + (x_F - 1) F_y = 18 \) (2) 3. \( x_F F_y - (y_F - 2) F_x = 40.5 \) (3) Now let’s manipulate these equations. From equation (1), we can express \( y_F F_x \): \[ y_F F_x = x_F F_y - 27 \quad (4) \] Substituting equation (4) into equation (2) gives: \[ -(x_F F_y - 27) + (x_F - 1) F_y = 18 \] This simplifies to: \[ -x_F F_y + 27 + (x_F - 1) F_y = 18 \implies (1 - 2)x_F F_y + 27 = 18 \implies -x_F F_y = -9 \implies x_F F_y = 9 \quad (5) \] Now, substituting equation (5) back into equation (4): \[ y_F F_x = 9 - 27 = -18 \quad (6) \] Next, substitute equations (5) and (6) into equation (3) to find \( F_x \) and \( F_y \): From equation (3): \[ 9 - (y_F - 2) F_x = 40.5 \] Using (6), we can substitute \( y_F = -\frac{18}{F_x} \): \[ 9 - \left(-\frac{18}{F_x} - 2\right) F_x = 40.5 \implies 9 + 18 + 2 F_x = 40.5 \implies 2F_x = 13.5 \implies F_x = 6.75 \] Now substituting \( F_x \) into (6) to get \( y_F \): \[ y_F F_{x} = -18 \implies y_F (6.75) = -18 \implies y_F = \frac{-18}{6.75} = -2.67 \] So the components of force \( \vec{F} \) are: \[ \vec{F} = (6.75, -2.67) \] Hence, the final result is: \[ \vec{F} = (6.75, -2.67) \, \text{units} \]