L₁ Rigid bar (1) Base E k D FIGURE P5.11

H WebAssign - ENGR 2331 Su2 x A 23 Changes in Lengths unde x yL.Het/web/Student/Assignment-Responses/last?dep326956848#Q3 : Apps M Gmail YouTube O Maps DETAILS UUMECHMATI Z.3.UT7. MT NOTES ASK TOUK TEACHEK A flat bar of rectangular cross section, length L, and constant thickness t is subjected to tension by forces P (see figure). b2 P b1 The width of the bar varies linearly from b, at the smaller end to b, at the larger end. Assume that the angle of taper is small. (a) Derive the following formula for the elongation of the bar. (Submit a file with a maximum size of 1 MB.) PL Et(b, - Choose File No file chosen This answer has not been graded yet. (b) Calculate the elongation (in inches), assuming L = 3 ft, t = 1.0 in., P = 23 kips, b, = 4.0 in., b, = 5.4 in., and E = 30 x 106 psi. (Use the deformation sign convention.) in. 73°F Partly sunny Watch it Need Help2 Read It O Type here to search hp

Topics Discussed: Static of Rigid Bodies, Force Vector, Addition of a System of Coplanar Forces, Cartesian Vector Please re-compute and re-solve the given, This is my answer. But I also want to see your expert answer if we land in the same solutions and answers. Please show your complete solution. Thank you.

Here I'm finding final kinetic energy KE2, given initial kinetic energy, initial force, alpha N/m^3 and beta N representing F(x). I integrated F(x)dx though I'm not sure how to integrate Fo with respect to X. I tried just finding the integrand ouput of 7.5x10^4m and subtracting the initial force, since the initial force would count as x=0, or I assumed. I then followed the regular approach of adding over the initial KE1 to isolate the unknown KE2 which I keeo getting as 1.07x10^10 which seems to be off. Again, I suspect where I'm going wrong is integrating the initial force -3.5x10^6N with respect to x, but other than what I've already done, that doesn't make sense to me.

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A rectangular tube of outer width w = 7.5 m, outer height h = 5 m and thickness t = 0.17 m experiences bending through the x axis. What is the moment of inertia of the cross-sectional area In? y -- X MATLAB input variables: format shortEng W = 7.5; h = 5; t = 0.17; moment of inertia I, = m4

Section 1.5 Problem 1. Mark each statement True or False, and justify your answer. (T/F) A homogeneous equation is always consistent. (T/F) If x is a nontrivial solution of Ax=0, then every entry in x is nonzero. (T/F) The equation Ax=0 gives an explicit description of its solution set. (T/F) The equation x = xu+xiv, with xa and xx free (and neither u nor va multiple of the other), describes a plane through the origin. (T/F) The homogeneous equation Ax=0 has the trivial solution if and only if the equation has at least one free variable. (T/F) The equation Ax=b is homogeneous if the zero vector is a solution. (T/F) The equation x = p + tv describes a line through v parallel to p. (T/F) The effect of adding p to a vector is to move the vector in a direction parallel to p.

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ools Fundamental of bio.. x 57 (73 of 286) e cdlaims compliance with the PDF/A standard and has been opened read-only to 4.13 Exercise Problems Problem 43 The uniform, horizontal beam shown in Fig. 4.51 is hinged to the wall at point A and supported by ar a cable attached to the beam at point C. Point C also th represents the center of gravity of the beam. At the other end, the cable is attached to the wall so that it makes an angle 0=68° with the horizontal. If the length of the beam is / = 4 m and the weight of the beam is W = 400 N, calculate the tension T in the cable and components of the reaction force on the beam at point A. Answers: T = 431 N, RAr = 162 N(→), RAy = 0 A Fie Fig. 4.51 Problem 4.3 Pr be 4.4 Using two different cable-pulley arrangements Problem

I'm wondering how z2 was derived. I've highlited where I am getting confused. Shouldn't the derivative of z2 should only be equal to x2; where, z2 = 0.5x^2/2. I don't understand how z2 = x2 multiplied by the derivative of x2.