# Frustrating Problem

by jrzygrl
Tags: frustrating
 P: 16 1. The problem statement, all variables and given/known data The well-known 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$$^{11}$$ E = (F / A) / ($$\Delta$$l / l) F = mg A = $$\pi$$r$$^{2}$$ 3. The attempt at a solution A = 4 $$\pi$$r$$^{2}$$ A = 4 $$\pi$$ (.01)$$^{2}$$ A = .00125m$$^{2}$$ F = mg F = 1030(9.8) F = 10094N l = 2.8m, $$\Delta$$l = .0046m E = (10094 / .00125) / (.0046 / 2.8) = 4.915 x 10$$^{11}$$ N/m$$^{2}$$ (2 x 10$$^{11}$$ - 4.915 x 10$$^{11}$$) = 1.951 x 10$$^{11}$$ 1.951 x 10$$^{11}$$ = (F / .00125) / (.0046 / 2.8) F = 400620.285 - 10094 = 390526.285N m = 39849.621kg The answer above was wrong, any help will be appreciated!
P: 413
hey, jrzygrl. Welcome to PhysicsForums.

 Quote by jrzygrl E = (10094 / .00125) / (.0046 / 2.8) = 4.915 x 10$$^{11}$$ N/m$$^{2}$$ (2 x 10$$^{11}$$ - 4.915 x 10$$^{11}$$) = 1.951 x 10$$^{11}$$
I don't get why you subtracted the Young's modulus [computed] for a given force from a given value. Even, if you did, you should've got the answer as zero, since, under the assumptions we're working under, the Young's modulus is constant for a given material. But it is not so in your case. This is because, in the equation, the force you've used is the force acting due to the walkway, but the strain you've used is NOT the strain caused by that force. The strain caused by the force will be such that, when put in the equation will give you the exact Young's modulus as the given data. What you've taken is the maximum strain the steel rods can tolerate and hence when put in the equation for Young's modulus, it is meaningless. To do so, you need a corresponding set of values for the Force and the strain it causes.

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:

$$F_i = \frac{10094 + x(9.8)}{4}$$

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, $F_i$, $A$, $\Delta l / l$ and the Young's modulus. I think you can take it from here :D
 P: 16 thanks a lot! i got it now

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