Question

Figure 2 of 3 Part C - Analyzing another force-couple system on a bracket The same bracket and force-couple system as that in Part B (F=159 lb, F = 53 lb, and M-590 lb-ft) is set up again, only this time, the resultant force is to be applied to the horizontal member CB.(Figure 2)Replace this force and couple system with an equivalent force by specifying the magnitude of the force, the direction in which the force acts, and where the force's line of action intersects member CB as measured from B. Express your answers, separated by commas, to three significant figures. The angle should be measured counterclockwise from the positive x axis (parallel to segment CB). View Available Hint(s) V A. vec moi ? Fr. Or, de lb, ft Submit

LUILurrelll luiLe system. ITIS 5 d Sysler where all the lines of action of the applied forces intersect at a common point Recause this means Part D-Analyzing a system of forces acting on a concrete slab igure 3 of 3 > The concrete slab shown in the picture is subject to four forces. Fi = 250 lb, Fy-365 lb, F3-680 lb, and F4 = 330 lb. (Figure 3) The dimensions are di = 8.5 ft, d2 = 18 ft, dz = 3.0 ft, ds = 14 ft, and ds = 10.0 ft. Determine the equivalent resultant force by specifying its magnitude and its location (2,y) on the slab. Express your answers, separated by commas, to three significant figures. View Available Hint(s) Y ALQ 11 vecm O ? FRI, Y= lb, ft, ft Submit

Answer 1

Analyzing a system of forces acting on a concrete slub: F = 250 lb ; F2 = 365lb ; F3 = 680l6; F = 330 16. dia 8.5 ft ; dq=18ft ; dz=3ft ; d4 = 14 ft ; ds = 10 ft. From tore balance, 4+ Frjo FR = Fit Fq + F + Fu Fr= 162516.] From momentum balance, .FR xC = F (ds) + Fq (dat ds) + Fu (dut ds) C = 600(10) + 365 (14+10) + 330 (14+10) 1625 C = 14.449 ft

- FR Y = F₂ Cd,) + Ficdit de) + F2 (d, tdqtd3) og - 0904.5) + 3900516)* 350 y : 680(8.5) + 250(8.5+1%) + 365 (8.571873) 1625 y = 13.798 st.