VI 1 where ag 9 80 m s2 VI 2 and VI 3 Measurements To calculate the force in Newtons Law you would use the equation F mkg ag Trial 1 Trial 2 F1 F2 F3 F1 F2 F3 F1 F2 F3 deg 112 198 349 115 200 323 118 222 328 F N 0 294 0 539 0 627 0 588 0 343 0 706 0 637 0 441 0 784 mt g 30 55 64 60 35 72 65 45 80 m1 mh 1000 We now need to convert grams to kilograms since the equation stands for the total mass Trial 3 exerted in kilograms So for trial 1 we take m1 from the grams and convert to kilograms mkg mh is already added into m1 as it is the mass of the hanger Looking at m1 we can use the equation mkg to find the kilograms m1 1000 30 1000 0 03 kg To calculate the force we use the equation F mkg ag F 1 mkg ag 0 03 kg 9 80 m s2 0 294 N VI 4 Measurements Trial 1 Trial 2 Trial 3 F1 F2 F3 F4 F1 F2 F3 F4 F1 F2 F3 F4 mt g 55 45 20 69 35 30 60 55 70 25 55 49 deg 0 80 160 229 10 90 170 297 20 100 180 262 F N 0 539 0 441 0 196 0 676 0 343 0 294 0 588 0 539 0 686 0 245 0 539 0 480 The calculations for F N are carried out the same way as in table 1 with converting grams to kilograms and then multiplying by 9 8 m s2 to find the force VI 2 Scale for both graphs 1cm 0 05 N Trial 3 for section VI 2 and VI 3 1 Force 1 0 637 N 1cm 0 05 N 12 74 cm 2 Force 2 0 441 N 1cm 0 05 N 8 82 cm 3 Force 3 0 784 N 1cm 0 05 N 15 68 cm Trial 1 for section VI 4 1 Force 1 0 539 N 1cm 0 05 N 10 78 cm 2 Force 2 0 441 N 1cm 0 05 N 8 82 cm 3 Force 3 0 196 N 1cm 0 05 N 3 92 cm 4 Force 4 0 676 N 1cm 0 05 N 13 52 cm VI 3 VI 4 The resultant is drawn on both diagrams It is way too small to obtain a measurement accurately The magnitude and direction of resultant is shown in VI 4 with some of the calculations shown To find the X component the equation is F x F cos Example calculation is as follows For the three force case F 1x 0 294 N cos 112 0 110 N The equation for F y is the same as F x except the trigonometric function is sin The equation for this would look like F y F sin An example calculation is as follows F y 0 294 N sin 112 0 273 N To find the x and y components of the resultant you would add all of the x components of the vectors and add all of the y components of the vectors For the x component calculation in trial 1 for the three forces 0 110 N 0 513 N 0 615 N 008N For the calculation of the y component in trial 1 for the three forces 0 273 N 0 176 N 0 120 N 0 023 N To find the magnitude of the resultant we need to use the equation C C x So calculation would be C 2 C y 2 To find the angle of the resultant you would take the x and y component of the resultant and use the equation tan 1 F y F x So calculation would be tan 1 0 023 0 008 70 82 3 Forces Data Table Trial 1 F1 F2 F3 0 110 0 513 0 615 0 273 0 167 0 120 xcomponent N y component N resultant x N resultanty N R N Degrees 008 0 023 0 024 70 82 x component N y component N resultant x N resultant y N R N Degrees x component N y component N resultantx N resultant y N R N Degrees 0 006 0 009 0 011 56 31 038 0 148 0 153 75 60 4 Forces Data Table Trial 2 F1 F2 F3 Trial 3 F1 F2 F3 Trial 1 F1 F2 F3 F4 Trial 2 F1 F2 F3 F4 Trial 3 F1 F2 F3 F4 0 248 0 322 0 564 0 299 0 328 0 665 N 0 539 0 0766 0 184 0 443 0 338 0 0 579 0 245 N 0 645 0 0425 0 539 0 0668 0 533 0 117 0 425 0 562 0 295 0 415 N 0 0 434 0 0670 0 510 0 0596 0 294 0 102 0 480 0 235 0 241 0 0 475 x component y component resultantx N resultant y N R N Degrees 0 0114 0 009 0 0145 38 29 xcomponent N ycomponent N resultant x N resultant y N R N Degrees 004 0 0244 0 0247 29 01 x component ycomponent N resultantx N resultant y N R N Degrees 0 0033 001 0 0034 5 16 86 All calculations are carried out the same way as they were for the three forces data The only difference is for the x and y components we need to add four vectors instead of three to find the resultant values So the new formula would be F R F1 F2 F3 F4 For x component F Rx 0 539 N 0 0766 N 0 184 N 0 433 N 0 0114 N The same process as shown above would be used for obtaining the value of the y component