When finding net force, why must you know the directions of the forces acting on an object? give two reasons

Atension test was conducted on a specimen of aisi 1020 hot-rolled steel having an initialdiameter of 9.11 mm. representative test data are given in table1in the form of force and engineering strain. for strain the extensometer gage length was li= 50.8 mm. in addition, minimumdiameters were measuredmanually with a micrometer in the necked region at several points during the test. after fracture, thebroken halves were reassembled, and the following measurements were made: (1) marks originally 25.4 mm apart and on opposite sides of the necked regionwere 38.6 mm apart due to the lengthwise stretching in the specimen; (2) similar marks originally 50.8 mm apart were 70.9 mmapart; (3) the final minimum diameter in the necked region was 5.28 mma) determine the following material properties: elastic modulus, 0.2% offset yield strength, ultimate tensile strength, percent elongation, and percent reduction in area. b) assume that the test was interrupted upon reaching a strain u = 0.007, and the specimen unloaded to zero force. estimate the elastic strain recovered and the plastic strain remaining. also, what would the new length of the original 50.8 mm gage section. hint: might save time/effort to use excel, mathcad, matlab, etc. to solve part a)table1 data and analysis for a tension test on aisi 1020 hot-rolled steeltest dataforce (p, kn)1. 0.002. 6.673. 13.344. 19.135. 17.796. 17.217. 17.538. 17.449. 17.2110. 20.7711. 24.2512. 25.7113. 25.7514. 25.0415. 23.4916. 21.3517. 18.9018. 17.39engr. strain (ε)1. 02. 0.00053. 0.00104. 0.00155. 0.00236. 0.00317. 0.00508. 0.00709. 0.010010. 0.049011. 0.125012. 0.218013. 0.234014. 0.306015. 0.330016. 0.348017. 0.360018. 03660diameter (d, mm)1. 9.112. 9.113. 9.114. 9.115. 9.116. 9.117. 9.118. 9.119. 9.1110. 8.8911. 8.8912. 8.2613. 8.2614. 7.6215. 6.9916. 6.3517. 5.7218. 5.28