Question

in case shown in (figure 1), determine the resultant couple moment at 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 b all attempts used; correct answer displayed (mr)o = value n·m

Explanation:

Step1: Resolve the force F into components

Let's assume the force \(F\) has a magnitude \(F\). Given the slope of the force vector as \(\frac{5}{13}\) for the vertical - to - hypotenuse ratio. If we consider the right - triangle formed by the components of the force, the vertical component \(F_y\) and horizontal component \(F_x\). Let's first find the vertical component of the force \(F\). The vertical component of the force \(F\) that creates a moment about point \(O\). The vertical component of the force \(F\) is \(F_y=\frac{12}{13}F\) and the horizontal component is \(F_x = \frac{5}{13}F\). Since the horizontal component of the force \(F\) passes through point \(O\), it does not create a moment about \(O\).

Step2: Calculate the moment due to each force about point \(O\)

The moment of a force \(M = r\times F\), where \(r\) is the perpendicular distance from the point of rotation to the line of action of the force.
The moment due to the \(200 - N\) force about \(O\) is \(M_1=200\times(2 + 2+2)=200\times6 = 1200\ N\cdot m\) (clock - wise, so negative).
The moment due to the \(400 - N\) force about \(O\) is \(M_2 = 400\times(2 + 2)=400\times4=1600\ N\cdot m\) (clock - wise, so negative).
Let's assume \(F = 1300\ N\) (since the slope is \(\frac{5}{13}\) and we can consider a convenient value for calculation), then \(F_y = 1200\ N\). The moment due to the vertical component of \(F\) about \(O\) is \(M_3=F_y\times2=1200\times2 = 2400\ N\cdot m\) (counter - clockwise, so positive).

Step3: Calculate the resultant couple moment about point \(O\)

\(M_R=(M_1 + M_2+M_3)\)
\(M_R=- 1200-1600 + 2400=-400\ N\cdot m\)

Answer:

\(-400\ N\cdot m\)