Question: When a flexible cable of uniform density is suspended between two fixed points and hangs of its own weight, the shape y = f(x) of
When a flexible cable of uniform density is suspended between two fixed points and hangs of its own weight, the shape y = f(x) of the cable must satisfy a differential equation of the form
-1.png)
Where is a positive constant. Consider the cable shown in the figure.
(a) let z = dy/dx in the differential equation. Solve the resulting first-order differential equation (in z), and then integrate to find y.
(b) Determine the length of the cable.
-2.png)
dx (b, h) (0, a) 0
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a d 2 ydx 2 k1 dydx 2 Setting z dydx we get dzdx k 1 z 2 dz1 z 2 k dx Using Formula 25 gives Where C ... View full answer
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