"FORCE, COMPRESSION and YOUNG's MODULUS. Below the elastic limit, or maximum stress that can be applied to an object without causing permanent deformation, a tension or compressive force FORCE actson an object of initial length LENGTH0 and cross-sectional area AREACROS to change its length by DELTALEN. Young's modulus YOUNGMOD is the ratio: (longitudinal stress)/(longitudinal strain) and it has the dimensions of a stress (N/m^2 or Pascal Pa and pound/in^2). *** Answer to problem *** (c) Copyright PCSCC, Inc., 1993 Note: the steel jack is assumed to have a circular cross section. Its area is thus pi*((outside diameter/2)^2-(inside diameter)^2). Steps to follow are: 1) Calculate cross sectional area of steel jack. Enter diameters in inches. 2) Determine the weight that produces the compression of length by solving Young's modulus equation for the force needed to yield DELTALEN. Set DELTALEN=.008, INSDIAM=2, LENGTH0=15*12 or 180 in, OUTSDIAM=3, and YOUNGSMO=3.5e7. The increased weight from the contraband is WEIGHT=6109 lbs. Type any key to exit. ||Agents are monitoring the weight of suspected contraband in a warehouse by monitoring the length decrease of a 15 ft. strategic hollow steel jack which supports the floor. (a) If the jack's inner diameter is 2 in., outerdiameter is 3 in., Young's modulus of the steel is 3.5*10^7 lb/in^2 and the tubelength decreased by .008 in, what is the weight (force) of contraband? Type comma key to see answers. Type (F2) to return to helpfile."
