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Frustrating Problemby jrzygrl
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#1
Apr508, 11:55 AM

P: 16

1. The problem statement, all variables and given/known data
The wellknown civil engineering firm of Rivers, Rhodes, and Waters is designing a walkway that is to be suspended by means of four steel rods of diameter 2.0 cm and length 2.80 m. The stretch of the rods is not to exceed 0.46 cm under any circumstances. What is the maximum additional mass that the 1030 kg walkway can carry to meet the design specifications? Assume that the load is uniformly distributed over the walkway so that each rod carries an equal share of the load. 2. Relevant equations Young's Modulus for steel (E) = 2 x 10[tex]^{11}[/tex] E = (F / A) / ([tex]\Delta[/tex]l / l) F = mg A = [tex]\pi[/tex]r[tex]^{2}[/tex] 3. The attempt at a solution A = 4 [tex]\pi[/tex]r[tex]^{2}[/tex] A = 4 [tex]\pi[/tex] (.01)[tex]^{2}[/tex] A = .00125m[tex]^{2}[/tex] F = mg F = 1030(9.8) F = 10094N l = 2.8m, [tex]\Delta[/tex]l = .0046m E = (10094 / .00125) / (.0046 / 2.8) = 4.915 x 10[tex]^{11}[/tex] N/m[tex]^{2}[/tex] (2 x 10[tex]^{11}[/tex]  4.915 x 10[tex]^{11}[/tex]) = 1.951 x 10[tex]^{11}[/tex] 1.951 x 10[tex]^{11}[/tex] = (F / .00125) / (.0046 / 2.8) F = 400620.285  10094 = 390526.285N m = 39849.621kg The answer above was wrong, any help will be appreciated! 


#2
Apr508, 12:10 PM

P: 413

hey, jrzygrl. Welcome to PhysicsForums.
To solve this problem, start by assuming the mass to be 'x'. Then, the force acting will be: F = 10094 + xg N Now, each of the rod can tolerate: [tex] F_i = \frac{10094 + x(9.8)}{4} [/tex] For a given rod, you have the force acting, 'F_i'. The force can atmost reach a point where it can cause a strain of 0.046/2.8 . So, we need to find the mass 'x', which would produce such a strain. This is the limiting mass as any mass greater than this would cause a higher strain, which is not permissible. Now, you have, [itex]F_i[/itex], [itex]A[/itex], [itex]\Delta l / l[/itex] and the Young's modulus. I think you can take it from here :D 


#3
Apr508, 06:33 PM

P: 16

thanks a lot! i got it now



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