G_-_2D_Equilibrium_of_Particles - 2-D EQUILIBRIUM OF A...

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Unformatted text preview: 2-D EQUILIBRIUM OF A PARTICLE PARTICLE LEARNING OBJECTIVES LEARNING Be able to draw a free body diagram (FBD). Be able to apply equations of equilibrium to solve a 2- D problem. PRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurements. Trigonometry concepts. Vector concepts. Rectangular components concepts. EQUILIBRIUM CONDITION OF A PARTICLE PARTICLE Newton’s second law of motion relates force and acceleration. A force operated on an object accelerates the object (it changes its velocity). The induced acceleration is proportional to the magnitude of the force and in the same direction as the force. The proportionality constant is the mass, m, of the object. ΣF = ma EQUILIBRIUM CONDITION OF A PARTICLE PARTICLE 1. A particle is at rest if it is originally at rest; v = 0, a = 0 2. A particle has a constant velocity if it is originally in motion; v = vo and a = 0 Newton’s Second Law: ΣF = ma, If the acceleration a = 0 then ΣF = m(0) = 0 and the object is in equilibrium COPLANAR FORCE EQUILIBRIUM EQUILIBRIUM ∑ F = 0 implies that ∑ F ⋅i = ∑ F ⋅ j = 0 x y The sum of the forces in the horizontal and vertical directions is equal to zero ∑ Fx = 0 ∑F y =0 FREE-BODY DIAGRAM (FBD) FREE-BODY A free-body diagram is a simple sketch that shows a free-body particle “free” from its surroundings with all the forces (both known and unknown) acting on it. (both Procedure for drawing a free-body diagram 1) Draw an outline showing the particle in question Draw isolated from its surroundings. isolated 1) Show all the active and reactive forces using arrows. 1) Identify all known forces by labeling their magnitudes Identify and directions. 1) Identify all unknown forces using letters. FBD-EXAMPLE FBD-EXAMPLE Draw FBD at knot A Y FAD A 30o FAB FAC X FBD-EXAMPLE FBD-EXAMPLE FBD at Knot CE FBD at Cord C FBD at the Ball Draw free-body diagrams for 1) The knot C 2) The cord CE 3) The ball SPRINGS SPRINGS The elastic characteristic of a linear spring is defined by The the spring constant or stiffness k. A force applied to a force linear spring would change its length by a distance S. The magnitude of the force is given by the following equation: equation: F = ks CABLES AND PULLEYS CABLES If a weightless cable cannot be stretched over a If frictionless pulley, the tension forces need to be constant in magnitude throughout the cable to keep the system in equilibrium. the T = Tension force only EXAMPLE EXAMPLE The spring ABC has a stiffness of 500 N/m and an at The rest length (unstretched length) of 6 m. Determine the magnitude of the horizontal force F that causes point B to move 1.5 m. causes SOLUTION SOLUTION Draw FBD at B: BC = 1.52 + 32 = 3.354 m B 1.5 m 3m FBA B C F Y FBC X SOLUTION SOLUTION The distance S & the magnitude of the force F in springs BA and BC are BC = 1.52 + 32 = 3.354 m B 1.5 m 3m S BA = S BC = BC terminal − BC initial = 3.354 − 3 = 0.354 m FBA = FBC = ks = (500)(0.354) = 177.1 N Equilibrium of horizontal forces: FBA B C F Y 1.5 ∑ Fx = (2)177.1 3.354 − F = 0 1.5 F = (2) 177.1 = 158.4 N 3.354 FBC X EXAMPLE EXAMPLE The cords BCA and CD The can each support a maximum load of 100 lb. Determine the maximum weight of the crate and the angle θ at equilibrium. equilibrium. SOLUTION SOLUTION Draw FBD at C Since CB and CA are of the same cord, the tension in each branch is equal to the weight of the crate W C Y W 13 5 12 T = 100 lb θ W X For equilibrium, the sum of the horizontal and vertical forces is equal to zero: 5 ∑ Fx = T cos θ − W 13 = 0 5 100 cos θ = W 13 12 ∑ Fy = T sin θ − W 13 − W = 0 25 100 sin θ = W 13 T =100 lb SOLUTION SOLUTION Cθ Y 13 5 12 W X W θ = 78.7o; W = 51 lb QUESTION When a particle is in equilibrium, the sum of forces acting on it is equal to ___ A) A constant B) A positive number C) Zero D) A negative number E) An integer. QUESTION A) A 100 lb F1 F2 40° A F2 40° 30° A 40° B) 30° F1 100 lb C) F 30° A 100 lb D) 30° A 100 lb Select the correct FBD of particle A QUESTION Y Using this FBD of Point C, the sum of forces in the y-direction (Σ Fy) is ___ . A) F2 sin 50° – 20 = 0 B) F2 cos 50° – 20 = 0 C) F2 sin 50° – F1 = 0 D) F2 cos 50° + 20 = 0 20 lb C F1 50° F2 X QUESTION For a frictionless pulley and cable, tensions in the cable (T1 and T2) are related as _____ . A) T1 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 sin θ T2 Cable in tension AA T1 ...
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