Question

replace the loading system (figure 1) by an equivalent resultant force and couple moment acting at point a. assume f1 = 150 lb, f2 = 250 lb, and f3 = 200 lb. part c determine the resultant couple moment acting at point a. express your answer to three significant figures and include the appropriate units. view available hint(s) (ma)r = value lb·ft incorrect; try again; 2 attempts remaining previous answers submit here we learn how to find the direction of the equivalent resultant force for a system of several forces and couple moments.

Explanation:

Step1: Calculate moment due to \(F_1\)

The perpendicular - distance from point \(A\) to the line of action of \(F_1\) is \(d_1=(3 + 3)=6\mathrm{ft}\). The moment due to \(F_1\) about point \(A\), \(M_1 = F_1\times d_1\). Given \(F_1 = 150\mathrm{lb}\), so \(M_1=150\times6 = 900\mathrm{lb - ft}\) (counter - clockwise).

Step2: Calculate moment due to \(F_2\)

The perpendicular - distance from point \(A\) to the line of action of \(F_2\) is \(d_2 = 4\mathrm{ft}\). The moment due to \(F_2\) about point \(A\), \(M_2=F_2\times d_2\). Given \(F_2 = 250\mathrm{lb}\), so \(M_2 = 250\times4=1000\mathrm{lb - ft}\) (counter - clockwise).

Step3: Calculate moment due to \(F_3\)

The perpendicular - distance from point \(A\) to the line of action of \(F_3\) is \(d_3 = 0\mathrm{ft}\) (since the line of action of \(F_3\) passes through point \(A\)), so \(M_3 = F_3\times d_3=200\times0 = 0\mathrm{lb - ft}\).

Step4: Calculate the resultant couple moment

The resultant couple moment \((M_A)_R\) about point \(A\) is the sum of the individual moments. \((M_A)_R=M_1 + M_2+M_3\). Substituting the values of \(M_1\), \(M_2\), and \(M_3\), we get \((M_A)_R=900 + 1000+0=1900\mathrm{lb - ft}\) (counter - clockwise).

Answer:

\(1900\mathrm{lb - ft}\)