TEGT3641 Engineering Mechanics II Dynamics UNIT 1B REVIEW OF STATIC EQUILIBRIUM 1 Revision of Equilibrium in Statics • Using a rectangular coordinate system, all forces exerted in the x, y and z directions must be balanced so that their sums equal to zero: z ΣFx = 0 o ΣFy = 0 y x ΣFz = 0 • All the moments must also be balanced: ΣMx = 0 ΣMo = 0 ΣMy = 0 } or ΣMz = 0 2 Static Equilibrium Using Free Body Diagram on the x-y Axis Force P applied on a stationery block of mass m Free Body Diagram W P m μ y P x F N The block is supported on a rough surface where μ is the coefficient of static friction; N is the Normal Force; F is the Friction Force and W is the Weight of the body whose Mass = m. 3 Solving for Static Equilibrium Force P applied on a stationery block of mass m Free Body Diagram W m P μ y x P F N Weight W = m.g Acceleration due to gravity g = 9.81 m/s2 Thus: W = 9.81.m where m is the mass (ii) Friction force F = μN (i) Note that the units of Force are Newton [N] 4 Solving for Static Equilibrium … Force P applied on a stationery block of mass m Free Body Diagram W m P μ y x P F N (iii) Conditions for Static Equilibrium →ΣFx = 0: or P – F = 0. Thus P = F and there is no motion in x-direction due to the friction. ↑ΣFy = 0: or N – W = 0. Thus N = W and there is no motion in the y-direction. 5 Statically Determinate Problems • An engineering problem based on static equilibrium with six unknowns can be solved ...
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