# Proj1 - /To μ for different values of To and a range of...

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Phys 125 Project 1 Problem: There is a chain of length L and mass M, suspended from two supports of equal height and separated by distance 2D. The tension between links of the chain is T(vector)=Txi+Tyj. Objective: The objective is to calculate the shape of the chain. Analysis: Consider a small piece of the chain with coordinates P1=(x1, y1) at the left end, and P2=(x1, y2) at the right end, with x2= x1 + x, and y2=y1+ y, then the length of that Δ Δ segment is s=[( x)^2+( y)^2)]^1/2, and its mass is ( s),where is the mass per unit Δ Δ Δ μ Δ μ length = M/L. The x-components of the tensions at points 1 and 2 have to be equal μ because the weight of the segment points in the vertical direction, but the vertical components differ by the weight g ( ). Call the horizontal component of the tension To. μ Δ We can plot the equation: u(x)=.5[exp(ax)-exp(-ax)]; a=g

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Unformatted text preview: /To μ for different values of To and a range of values of x: x=-0.4 : 0.01 : 0.4; t=5; u=.5*[exp((29.4*x)/t)-exp((-29.4*x)/t)]; z=(0*x)*(14.7/t); plot(x,u,x,z); this display of matlab commands is shown in Figure 1. Figure 1 shows what xmax is. By making t=100 you get Figure 2 and by making t=1000 you get Figure 3. If we integrate u(x) we get y(x): y(x)= (To/58.8)(exp((29.4x)/To)+exp((-29.4x)/To)) for different values of To we can plot how the chain would look: x= -.4: .01 : 0.4; t= 5; y= .5*[exp((29.4*x)/t)-exp((-29.4*x)/t)]; plot(x,y) this display of matlab is shown in Figure 4. Figure 4, Figure 5, and Figure 6 show how the chain looks for different values of To. Conclusion: Increases in To will also increase the maximum x tension in each chain link. Also increases in To will increase the amount of sag in the chain....
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Proj1 - /To μ for different values of To and a range of...

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