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Additional resources for Dynamic system modeling and control
1 Ns -----m ans. 6 Cables And Pulleys Cables are useful when transmitting tensile forces or displacements. The centerline of the cable becomes the centerline for the force. And, if the force becomes compressive, the cable becomes limp, and will not transmit force. A cable by itself can be represented as a force vector. When used in combination with pulleys, a cable can redirect a force vector or multiply a force. Typically we assume that a pulley is massless and frictionless (in the rotation chapter we will assume they are not).
In cases where there is no lubricant, and the touching surfaces are dry, dry coulomb friction may result. In this case the surfaces will stick in place until a maximum force is overcome. After that the object will begin to slide and a constant friction force will result. 26 shows the classic model for (dry Coulomb) friction. The force on the horizontal axis is the force applied to the friction surfaces while the vertical axis is the resulting friction force. Beneath the slip force the object will stay in place.
After this, the equation is rearranged into Hooke’s law, and the equivalent spring coefficient is found. 33 First, draw FBDs for P and M and sum the forces assuming the system is static. K S1 y 2 P + ∑ F y = K S1 y 2 – K S2 ( y 1 – y 2 ) = 0 (1) K S2 ( y 1 – y 2 ) K S1 K S2 ( y 1 – y 2 ) K S2 M + ∑ Fy P = K S2 ( y 1 – y 2 ) – F g = 0 (2) Fg M y1 Next, rearrange the equations to eliminate y2 and simplify. (1) becomes K S2 ( y 1 – y 2 ) – Fg = 0 Fg y 1 – y 2 = --------K S2 Fg y 2 = y 1 – ---------K S2 (2) becomes (3) K S1 y 2 – K S2 ( y 1 – y 2 ) = 0 (4) y 2 ( K S1 + K S2 ) = y 1 K S2 Fg sub (3) into (4) y – --------- ( K + K ) = y K S2 1 S2 1 K S2 S1 Fg K S2 y 1 – ---------- = y 1 ---------------------------K S2 K S1 + K S2 K S2 F g = y 1 1 – ---------------------------- K S2 K S1 + K S2 K S1 + K S2 – K S2 F g = y 1 -------------------------------------------- K S2 K S1 + K S2 K S1 K S2 F g = y 1 ---------------------------- K S1 + K S2 Finally, consider the basic spring equation to find the equivalent spring coefficient.