Chapter 4, Problem 32FP

Find the resultant of the concurrent forces R and S for which the direction is specified by slope, expressed as rise and run values. Show a sketch of R and S and the resultant   R= 8.6(10)^3 N Rise= -3.0 Run= 4.0 S= 6.2(10)^3 N Rise= 3.0 Run= 4.0

Using the Component method, determine the magnitude and direction of the resultant and the equilibrant of the given forces.                   Force A = 316lb / 20° E of S                  Force B = 518lb / 15° N of E                  Force C = 111lb / W 05° S

Engineering Mechanics: Statics

PROBLEM 1. Replace the loading system by an equivalent resultant force and specify where the resultant's line of action intersects the member AB measured from A. 0.5 m -1.5 m- 0.5 m 0.5 m| B 8 kN 6 kN 5 kN 3 m A

The diagram shows the distributed loading profile on a vertical surface. Replace the distributed load by a single force and determine its location along the surface with respect to point O. Q4 3 m 7.5 kN/m 3 m 4.5 m 20 kN/m

Determine the magnitude of the resultant force F(R). Determine the angle between the resultant force and the x axis. And finally specify where the resultant’s line of action intersects member AB, measured from point A. (y = ….. ft above A)

Replace the three forces acting on the bent pipe by a single equivalent force R. Specify the distance d from point O to the point on the x-axis through which the line of action of R passes: 50 lb 40 lb 10 10" 10" 60 lb

Determine the resultant R of the system of distributed loads and locate its line of action with respect to the left support for y 400 Ib/ft 300 Ib/ft 150 Ib/ft A 7 ft 8 ft

Replace the distributed loading shown with an equivalent resultant force, calculating both (a) the magnitude of the resultant force and (b) the location of the resultant force, as measured from point B.