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Unformatted text preview: 2-D EQUILIBRIUM OF A PARTICLE LEARNING OBJECTIVES 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 Units of measurements. Trigonometry concepts. Vector concepts. Rectangular components concepts. EQUILIBRIUM CONDITION OF A 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. F = ma EQUILIBRIUM CONDITION OF A 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: F = ma, If the acceleration a = 0 then F = m(0) = 0 and the object is in equilibrium COPLANAR FORCE 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)
A free-body diagram is a simple sketch that shows a particle "free" from its surroundings with all the forces (both known and unknown) acting on it. Procedure for drawing a free-body diagram
1) Draw an outline showing the particle in question isolated from its surroundings.
2) Show all the active and reactive forces using arrows. 3) Identify all known forces by labeling their magnitudes and directions.
4) Identify all unknown forces using letters. FBD-EXAMPLE
Draw FBD at knot A Y
30o FAB FAC X 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
The elastic characteristic of a linear spring is defined by the spring constant or stiffness k. A force applied to a linear spring would change its length by a distance S. The magnitude of the force is given by the following equation: F = ks CABLES AND PULLEYS
If a weightless cable cannot be stretched over a frictionless pulley, the tension forces need to be constant in magnitude throughout the cable to keep the system in equilibrium. T = Tension force only EXAMPLE The spring ABC has a stiffness of 500 N/m and an at rest length (unstretched length) of 6 m. Determine the magnitude of the horizontal force F that causes point B to move 1.5 m. 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
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
The cords BCA and CD can each support a maximum load of 100 lb. Determine the maximum weight of the crate and the angle at equilibrium. 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 W X For equilibrium, the sum of the horizontal and vertical forces is equal to zero:
5 Fx = T cos - W 13 = 0 5 100 cos = W 13 12 Fy = T sin - W 13 - W = 0 25 100 sin = W 13 T =100 lb SOLUTION C Y
13 5 12 W X W = 78.7o; W = 51 lb ...
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- Summer '08