Physics · Ch 4 — Laws of Motion
Mechanical Equilibrium
Mechanical Equilibrium
As a direct consequence of Newton's second law, the momentum of a system stays constant whenever no external unbalanced force acts on it -- this state is called MECHANICAL EQUILIBRIUM. A particle is in mechanical equilibrium if the net force on it is zero; a SYSTEM of bodies is in mechanical equilibrium if the net force on every individual part of the system is zero, which requires the velocity (or linear momentum) of every part to stay constant (possibly zero), with no acceleration anywhere in the system. Mathematically, for every part of the system, for it to be in mechanical equilibrium. In everyday usage the word 'couple' is sometimes used loosely to mean the same thing as 'the moment/torque of a couple' -- i.e. it is common to simply say 'a couple is acting' rather than always explicitly saying 'a torque due to a couple is acting'.
Worked illustration (plank on two cables, Example 4.10): a 30 kg plank hangs symmetrically from two vertical cables 2 m apart, each rated for a maximum tension of 500 N; a 50 kg boy standing at the centre wants to know how far he can walk towards one cable. Taking torques about the far support point A, with the plank's weight (300 N, acting 1 m from either cable) and the boy's weight (500 N, at distance x from centre) both producing a clockwise moment balanced by the anticlockwise moment of the maximum tension in the near cable (500 N at 2 m from A), solving gives m -- the boy can walk up to 40 cm to either side of the centre before exceeding a cable's rated tension. …
Worked out. A 30 kg wooden plank is supported symmetrically by two vertical cables 2 m apart, each rated to sustain up to 500 N tension; a 50 kg boy standing at the plank's centre wants to know how far towards one cable he can walk. Taking moments about the far (left) support A, with the weight of the plank (300 N, 1 m from either cable) and the boy's weight (500 N, at distance x from centre) both producing clockwise moments balanced by the anticlockwise moment of the maximum tension in the near cable (500 N, at 2 m from A), the example solves 300(1) + 500(1+x) = 500(2) to get x = 0.4 m, i.e., the boy can walk up to 40 cm on either side of the centre before a …
Worked out. A negligible-mass ladder with a horizontal cross-bar (like a stepladder) rests on a frictionless floor, its two 1 m legs spread at a 40-degree angle between them, with a 50 kg person standing at the top where the legs meet; because the floor is frictionless, the normal reaction at each foot is purely vertical, and by symmetry each foot carries half the person's weight, N = mg/2 = 250 N. Taking torques about the top of one leg, the clockwise torque of N (acting at the base) must balance the anticlockwise torque of the horizontal cross-bar tension T (acting partway up the leg), from which the tension T in the cross-bar is cal …