span it follows from the argument in section 21 that

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Unformatted text preview: e previous section. For example, let us consider the mechanical system illustrated in Figure 2.5. Here we have two carts moving along a friction-free track, each containing mass m and attached together by three springs, with spring constants k1 , k2 and k3 . Let x1 (t) x2 (t) F1 F2 = the position of the first cart to the right of equilibrium, = the position of the second cart to the right of equilibrium, = force acting on the first cart, = force acting on the second cart, with positive values for F1 or F2 indicating that the forces are pulling to the right, negative values that the forces are pulling to the left. Suppose that when the carts are in equilibrium, the springs are also in equilibrium and exert no forces on the carts. In this simple case, it is possible to reason directly from Hooke’s law that the forces F1 and F2 must be given by the formulae F1 = −k1 x1 + k2 (x2 − x1 ), F2 = k2 (x1 − x2 ) − k3 x2 , but it becomes difficult to determine the forces for more complicated mechanical systems consisting of many weights and springs. Fortunately, there are some simple principles from physics which simply t...
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This document was uploaded on 01/12/2014.

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