Question

Suppose you are going to measure Young's modulus for three rods by measuring their stretch when they are suspended vertically and weights are hung from them, as shown. (FIA) Rod 1 is 3.7 meters long and cylindrical with radius 7 mm (1 millimeter is 0.001 m). Rod 2 is 2.2 meters long by 8 mm wide by 3 mm deep. Rod 3 is 2.0 meters long by 6 mm wide by 6 mm deep. The definition of Young's modulus, Y = s, includes the quantity A, the cross-sectional area. (ALIL)' *Part 1 (a) What is the cross-sectional area of rod 1? Aj =U m2 the tolerance is +/-5% Attempts: 0 of 10 used *Part 2 (b) What is the cross-sectional area of rod 2? A2 = 4 m2 the tolerance is +/-5% Attempts: 0 of 10 used *Part 3 (c) What is the cross-sectional area of rod 3? Az =U * m2 the tolerance is +/-5% Attempts: 0 of 10 used

Answer 1

- Given length of rad 1 (cylindrical) 1 = 3.7m - radius of rod 2 (8)= = 7mm. 0.007.m length of rod 2 wideness of rod 2 deepness of rod 2 Ccuboid) - 2.2m = 8mm = 0.008 m = mm - 0.oozm length wideness deepness of of of rod 3 rad 3 rod 3 (cuboid ) = 2.om = 6mm = 0.006 m. - 6mm = 0.006 m (FTA module = formula of Young's (A/L) of rod 1 - (9) Cross section area A = A (radius of yod 192 A = 22 x (0.007)2 m² A = 22x7x100 m2 7

(b) Cross section area of roda- 4 = (wideness X deepnes) op rod 2 A f0.008 m) C0.003m) [Ą - 2.4x105 m² (C) Cross section area of rad - Az = (wideness x deepness) Op rod 3 A = (0.006x0.006) m² 2/ E 3.6 x 10-5 m2