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Chapter 2. Coplanar Concurrent Forces. Introduction. In the first chapter we discussed about simple case of concurrent forces, where only two non parallel forces were considered.
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Chapter 2 Coplanar Concurrent Forces
Introduction • In the first chapter we discussed about simple case of concurrent forces, where only two non parallel forces were considered. • However, in many cases, the lines of action of three or more forces may intersect at a single point, and can form a system of concurrent forces. • In this chapter we are going to discuss these type of coplanar concurrent forces.
Resultant of Concurrent Force (Algebraically) • Find ∑ Fx, ∑ Fy then determine the resultant force and the angle. (The sense in which each force acts is considered) • Ax = A cos α, Bx = B cos β, Cx = C cos γ • Ay = A sin α, By = B sin β, Cy = C sin γ • ∑ Fx = A cos α - B cos β - C cos γ • ∑ Fy = A sin α - B sin β - C sin γ • Resultant force, R = [(∑ Fx)2 + (∑ Fy)2]1/2 • Direction or angle, tanθ′ = ∑ Fy / ∑ Fx
Example- 16 • Find algebraically the resultant of the force system shown in Fig 20 (a)
Equilibrium • A system of forces where the resultant is zero is said to be in Equilibrium. (Defined by Newton’s First Law) • Now If, R= 0 then, • ∑ Fx = 0 • ∑ Fy = 0 • The above two equations can be used to find two unknowns.
Problem 57 • R is the resultant of F, T and Q, shown in Fig. 33. F = 150 lb., θ = 30º, R = 85 lb. Find Q and α. • Example 21- Do it yourself
Free Body • Most bodies in equilibrium are at rest. • But a rigid body moving with constant speed in a straight path is also in equilibrium. • A rigid body may be any particular mass whose shape remains unchanged while it is being analyzed for the effect of forces. • Since no body is truly rigid, we mean by ‘rigid body’ is one whose deformation under force is negligible for the purposes of the problem. • A free body is a representation of an object, usually a rigid body , which shows all the forces acting on it. (See page 22-23 of Analytic Mechanics by Virgil Moring Faires for details)
Problem 72 • A 5000 lb sphere rests on a smooth plane inclined at an angle θ = 45 º with the horizontal and against a smooth vertical wall. What are reactions at the contact surface A and B, Fig. 39?
Equilibrium of Three Forces & Force Polygon Again from the equation of equilibrium, • ∑Fx = F2 cos45º - F1 cos60º = 0 • ∑Fy = F2 sin45º + F1 sin60º = 0
Problem: 91 • Two weights are suspended from a flexible cable as shown in Fig. 46. For θ = 120º, determine the internal forces in various parts of the cable and weight W.
Trusses ? • Trusses are mainly used to support Roofs and Bridges. Roof Truss Bridge Truss
Trusses: Joint to joint method • A determinate structure is one wherein the internal forces in the various members of the structure may be obtained by the conditions of equilibrium. • Although truss is subjected to all manner of loading, for simplicity it is assumed that, the loads on truss are applied at pin joints • So, with a truss loaded in this manner, all the various members are two force members and the free body of each pin is a system of concurrent forces. • If all the external forces including the reactions at the supports are known, then the above principles can be used to determine the internal forces in each member. • To determine the member forces, it is assumed that the unknown forces are acting away from the pin, which means that the members are in tension. • Solve for each of the unknowns by using the conditions of equilibrium, i.e., ∑ Fx = 0, ∑ Fy = 0.
Example 22 • A roof truss is constructed and loaded as shown in Fig. 27. At the pin N, the following internal forces have been found: NF = 16900 lb (tension), DN = 7000 lb (compression) and NH = 12565 lb (tension). For an external load at the pin of 6000, find the forces in the members NC and NB.
Closure • See Book Page: 32 from Analytic Mechanics by Virgil Moring Faires
Assignments From Book (Analytic Mechanics by Virgil Moring Faires): • Problem: 77 • Problem: 79 • Problem: 114 • Problem: 137 • Problem: 138
