Problem 2: A force \( P \) in the \( x y- \) plane acts on the triangular plate. The moments of \( P \) about points \( O, A \), and \( B \) are \( M_{o}=80 \mathrm{~N} \cdot m \) (counterclockwise), \( M_{A}=200 \mathrm{~N} \cdot \mathrm{m} \) (clockwise), and \( M_{B}=0 \). Determine \( P \).

To determine the force \( P \) acting on the triangular plate, we need to analyze the moments about the points \( O \), \( A \), and \( B \). ### Given: - Moment about point \( O \): \( M_O = 80 \, \text{N} \cdot \text{m} \) (counterclockwise) - Moment about point \( A \): \( M_A = 200 \, \text{N} \cdot \text{m} \) (clockwise) - Moment about point \( B \): \( M_B = 0 \) ### Step 1: Moment about point \( O \) The moment \( M_O \) is given by the equation: \[ M_O = r_O \cdot P \cdot \sin(\theta) \] where \( r_O \) is the distance from point \( O \) to the line of action of force \( P \), and \( \theta \) is the angle between the line connecting point \( O \) to the point of application of \( P \) and the direction of \( P \). Since \( M_O = 80 \, \text{N} \cdot \text{m} \) (counterclockwise), we can write: \[ 80 = r_O \cdot P \cdot \sin(\theta) \tag{1} \] ### Step 2: Moment about point \( A \) The moment \( M_A \) is given by: \[ M_A = r_A \cdot P \cdot \sin(\theta) \] where \( r_A \) is the distance from point \( A \) to the line of action of force \( P \). Since \( M_A = 200 \, \text{N} \cdot \text{m} \) (clockwise), we can write: \[ 200 = r_A \cdot P \cdot \sin(\theta) \tag{2} \] ### Step 3: Moment about point \( B \) The moment \( M_B \) is given by: \[ M_B = r_B \cdot P \cdot \sin(\theta) \] where \( r_B \) is the distance from point \( B \) to the line of action of force \( P \). Since \( M_B = 0 \), we can write: \[ 0 = r_B \cdot P \cdot \sin(\theta) \tag{3} \] ### Step 4: Analyzing the equations From equation (3), since \( M_B = 0 \), either \( P = 0 \) or \( r_B = 0 \) or \( \sin(\theta) = 0 \). However, since \( P \) is a force acting on the plate, we can assume \( P \neq 0 \) and \( \sin(\theta) \neq 0 \). Therefore, we conclude that \( r_B = 0 \), meaning that point \( B \) is on the line of action of force \( P \). ### Step 5: Relating the distances Since \( r_B = 0 \), we can relate \( r_O \) and \( r_A \) to the distances from points \( O \) and \( A \) to point \( B \). Let’s denote: - \( d_O \) as the distance from \( O \) to \( B \) - \( d_A \) as the distance from \( A \) to \( B \) Then we can express: \[ r_O = d_O \quad \text{and} \quad r_A = d_A \] ### Step 6: Solving for \( P \) From equations (1) and (2), we can express \( P \): \[ P = \frac{80}{d_O \cdot \sin(\theta)} \tag{4} \] \[ P = \frac{200}{d_A \cdot \sin(\theta)} \tag{5} \] Setting equations (4) and (5) equal to each other: \[ \frac{80}{d_O \cdot \sin(\theta)} = \frac{200}{d_A \cdot \sin(\theta)} \] Cancelling \( \sin(\theta) \) (assuming it is not zero): \[ 80 \cdot d_A = 200 \cdot d_O \] \[ d_A = \frac{200}{80} \cdot d_O = 2.5 \cdot d_O \] ### Step 7: Substituting back to find \( P \) Now substituting \( d_A = 2.5 \cd