Equilibrium of Concurrent, Coplanar Force Systems

Equilibrium of Concurrent, Coplanar Force Systems

Equilibrium of Concurrent, Coplanar Force Systems EF 202 - Week 5 1 Equilibrium Newtons First Law - If, and only if, an objects mass center has zero acceleration, then the sum

of ALL of the forces acting on the object (body) is zero. Where is a bodys mass center? Well study that in Module 4. But in this class, the entire body is at rest, so we know that the mass center, wherever it is, has zero acceleration. 2 Problem Solving

Strategy 1.Identify ALL forces acting on the body by making a free-body diagram (FBD). 2.Invoke Newtons First Law. (Add all the force vectors on the FBD and set the result equal to zero.) 3.Solve. 3 Free-Body Diagram 1.Define the body.

2.Free the body from the rest of the world by cutting through EVERYTHING that connects the body to anything else. 3.Sketch the freed body. 4.To the sketch, add EVERY force that is (or may be) acting on the body. 4 Concurrent, Coplanar Force System Coplanar system

The lines of action of all forces lie in a common plane. Concurrent system The lines of action of all forces intersect at a common point. 5 Independent Equations In a given problem, we can

find only as many unknowns as we have independent equations. For a system of coplanar, concurrent forces, Newtons First Law yields only two independent equations for a given FBD. 6 Problem 3-4 Find magnitude and

direction of F. 1. Make FBD. 2. Newtons First Law 3. Solve. 1. Figure is already an FBD.r F =0 2. F (cos i + sinj) + (4.5 kN) i + L o

L 7.5 kN(sin 30 i cos 30 j) + L o o L 2.25 kN(cos60 i sin60 j) = 0 o 7 Collect and equate components. For the i components: o

o F cos 4.5 kN 7.5 kN sin 30 + 2.25 kN cos60 = 0 (1) the j components: For F sin 7.5 kN cos 30 o 2.25 kN sin60 o = 0 (2) Now, what?

COUNT EQUATIONS AND UNKNOWNS!!!!!!! Two equations, (1) and (2), and two unknowns, F and . 8 From (1) and (2), F cos = 7.125 kN (3)

F sin = 8.444 kN (4) Now, what? For the angle, divide (4) by (3) to get tan = 1.185 = 49.8 o For the magnitude,

2 x 2 2 y F = F + F = ( F cos ) + ( F sin ) F = 11.0 kN 9 2

Ideal Cable Neglect weight (massless). Neglect bending stiffness. Force parallel to cable. Force only tensile. Neglect stretching (inextensible). 1 0 Problem 3-8 Find the forces in cables AB and AC. 1. Make FBD.

2. Newtons First Law 12 kg 3. Solve. 1 1 COUNT EQUATIONS AND UNKNOWNS!!!!!!! 1 2 1

3 1 4 Problem 3-40 Find the forces in all five cables. Since there are 5 unknowns, well need at least 3 FBDs. 30kg 1

5 1 6 1 7 Linear Spring I Like an ideal cable, but extensible and resists either stretching or shortening. Force is either tensile or

compressive. When the force is zero, the spring is undeformed and its length is called its free length or natural length. When the force is not zero, it is proportional to the springs deformation (stretch or contraction). 1 8 Linear Spring II

The constant of proportionality between the force in a spring and its deformation is called the springs stiffness or its spring rate. Stiffness is usually represented by the letter k.

1 9 Equation for Spring Tensio n Compression OR F = k ( l l0 )

123 F = k ( l0 l) 123 stretch contraction l0 = natural/free/undeformed length (length when F = 0)

2 0 Assumed Tension F = k ( l l0 ) Directions of loads are reversed on objects to which springs are attached (Newton III). 2

1 Assumed Compression F = k ( l0 l) Direction assumed for spring load cannot be wrong! Equation

used for spring load can be wrong! 2 2 Problem 3-14 Spring AB is stretched 3m. What is the mass of the suspended block? 2 3

2 4 2 5

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