ch 3 statics of particles
TRANSCRIPT
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CH 3STATICS OF PARTICLES
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System ofForces
Coplanar 2D
Concurrent Nonconcurrent
ParallelGeneral
Noncoplanar 3D
Nonconcurrent
Parallel General
Concurrent
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Free-Body Diagrams
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Free-Body Diagrams
Space Diagram: A sketch showing
the physical conditions of the
problem.
Free-Body Diagram: A sketch showing
only the forces on the selected particle.
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Equilibrium of a Particle• When the resultant of all forces acting on a particle is zero, the particle is
in equilibrium.
• Particle acted upon by
two forces:
- equal magnitude
- same line of action
- opposite sense
• Particle acted upon by three or more forces:
- graphical solution yields a closed polygon
- algebraic solution
00
0
==
==
yx FF
FR
• Newton’s First Law: If the resultant force on a particle is zero, the particle will
remain at rest or will continue at constant speed in a straight line.
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example
◼ The cable system shown in Fig. 3-8 is being
used to lift body A. The cable system is in
equilibrium at the cable positions shown in the
figure when a 500-N force is applied at joint 1.
Determine the tensions in cables A, B, C, and
D and the mass of a body A that is being lifted.
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◼ 2.10 To steady a sign as it is being lowered, two cables are
attached to the sign at A. Using trigonometry and knowing that
the magnitude of P is 300 N, determine
a) The required angle if the responding R of the two forces
applied at A is to be vertical
b) The corresponding magnitude of R.
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Solution (a)
Using the triangle rule and the law of sines
360 N
300 N
35˚
R
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Solution (b)
β= ?We need to calculate the value of βto find the value
of R
Law of Sinus a/sin A = b/sin B = c/sin C
R = 513 N
360 N
300 N
35˚
R
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Equilibrium of Forces
F4 = 400 N
F3 = 200 N
F2 = 173.2 F1 = 300
O
OF1 = 300
F2 = 173.2 N F3 = 200
F4 = 400 N
30
30
An equilibrium system of forces
produces a closed force polygon
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Example Figure .27 shows four forces acting on A.In figure .28, the resultant of the given forcesis determined by the polygon rule. Starting from point O with F1 and arranging the forces in tip-to-tail fashion,we find that the tip of F4 coincides with thestarting point O. Thus the resultant R of the given system of forces is zero, and the particle is in equilibrium.
The closed polygon drawn in Fig 2.28 provides a graphical expression of the equilibrium of A.To express algebraically the condition for the equilibrium of a particle, we write
Resolving each force F into rectangular components,we have
We conclude that the necessary and sufficientconditions for the equilibrium of a particle are
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Returning to the particle shown in figure 2.27, we check that the equilibrium conditions are satisfied. We write