Question

replace the loading shown in (figure 1) by an equivalent resultant force and couple moment acting at point a. suppose that w = 350 n/m. part a determine the equivalent resultant force. express your answer to three significant figures and include the appropriate units. assume the positive direction is upward. part b determine the couple moment. assume the positive direction is counterclockwise.

Explanation:

Step1: Calculate the resultant force of the distributed - load

The distributed - load can be divided into a rectangular part and a triangular part. The rectangular part has a load intensity of \(w = 350\ N/m\) and a length \(L_1=3\ m\), and the triangular part has a base length \(L_2 = 3\ m\) and a maximum load intensity of \(w = 350\ N/m\).
The force due to the rectangular part \(F_1=w\times L_1=350\times3 = 1050\ N\).
The force due to the triangular part \(F_2=\frac{1}{2}\times w\times L_2=\frac{1}{2}\times350\times3 = 525\ N\).
The resultant force \(F_R=F_1 + F_2\).
\[F_R=1050+525=1575\ N\]

Step2: Calculate the couple - moment about point A

The rectangular part of the load creates a moment about point A. The centroid of the rectangular part is at \(x_1=\frac{3}{2}=1.5\ m\) from point A. The moment due to the rectangular part \(M_1=F_1\times x_1=1050\times1.5 = 1575\ N\cdot m\).
The centroid of the triangular part is at \(x_2 = 3+\frac{1}{3}\times3=4\ m\) from point A. The moment due to the triangular part \(M_2=F_2\times x_2=525\times4 = 2100\ N\cdot m\).
The couple - moment \(M = M_1+M_2\).
\[M=1575 + 2100=3675\ N\cdot m\]

Answer:

Part A:
\(F_R = 1575\ N\)
Part B:
\(M = 3675\ N\cdot m\)