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.

Explanation:

Step1: Analyze the loading

The loading consists of a uniformly - distributed load (UDL) over the upper 3 m and a triangular - distributed load (TDL) over the lower 3 m.

Step2: Calculate the resultant of UDL

The magnitude of the uniformly - distributed load \(w = 350\ N/m\) and the length \(L_1=3\ m\). The resultant of the UDL, \(F_{1}=w\times L_1\). So, \(F_{1}=350\times3 = 1050\ N\), and it acts at the mid - point of the UDL, i.e., \(x_1 = 1.5\ m\) from point \(A\).

Step3: Calculate the resultant of TDL

The triangular - distributed load has a maximum intensity of \(w = 350\ N/m\) and length \(L_2 = 3\ m\). The magnitude of the resultant of the TDL, \(F_{2}=\frac{1}{2}\times w\times L_2=\frac{1}{2}\times350\times3= 525\ N\). It acts at a distance \(x_2=\frac{1}{3}\times3 = 1\ m\) from the base of the triangle (from point \(B\)), so \(x_2 = 3 + 1=4\ m\) from point \(A\).

Step4: Calculate the total resultant force

The total resultant force \(F_R=F_1 + F_2\). Substituting the values, \(F_R=1050+525=1575\ N\).

Answer:

\(F_R = 1580\ N\) (rounded to three significant figures)