Assignment 2 Outline

# In general we will have points xn1 yn1 and xn yn

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Unformatted text preview: a To find the length of the chain for a given a, break up the chain into a sequence of straight- line segments, and add up the lengths of the segments. In Figure 2, the chain has been broken into five segments using six equally- spaced x values: x0, x1, x2, x3, x4, x5., where x0 = –d and x5 = d. The corresponding y values, y0, y1, y2, y3, y4, y5, are found from Equation (1) using a ~ 0.7 [m] for this example. To get an accurate estimate of the length, many more segments than five are required. The length of a single segment between (xj–1, yj–1) and (xj, yj) is ( ) +( ) [m]. Given X = {xj}j=0,n and Y = {yj}j=0,n, the general formula for an estimate of the length of the entire chain in metres using n segments is 2 cosh(x) = (exp(x) + exp(- x)) / 2 COMP 1012 Winter 2014 Assignment 2 L( , )= (2) ( Page 3 of 9 ) +( ) = (x0, y0) (x5, y5) (x1, y1) (x4, y4) (x2, y2) (x3, y3) Figure 2: Line Segment Approximation Finding the Force on a Support Consider the last segment in the approximation to a continuous chain, as shown in Figure 2, and close up in Figure 3. In general we will have points (xn–1, yn–1) and (xn, yn) showing. Fnet θ Fvert θ Fhor Figure 3: Last Segment The net force on the support will be a vector parallel to this segment, pointing down and to the left. By symmetry, the force vertically downward on each support, Fvert, will represent half the chain’s weight. If the chain has density ρ [kg/m], then (3) | |= where g [m/s2] is the accele...
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## This note was uploaded on 02/10/2014 for the course COMP 1012 taught by Professor Terryandres during the Winter '14 term at Manitoba.

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