Question

Determine equivalent force-couple system at A. If Fac-999 N and F-1998 N write both resultant and moment about A in form of sum of components (xi +y k) Identify coordinates of points A ( в ale D ( E( 3 mi 3 un 1,5 m Set up and solve equations for sums of the forces in each direction a. i + b. Fill in the blanks: R = Set up matrices needed to determine moment about A c. d. Resolve matrices in part (c). Write your answer in the form M = k i+ i +

Answer 1

EQUIVALENT FORCE-COUPLE SYSTEM AT A along B D Force F The force is directed from whose coordinates are as follows: B to D в( 0 0) m 3 3 m 0.0 D 1.5 Therefore, the position vectors are: 0 i 3 j OB= O D 1.5 + 3 k 0 O D - BD = Hence, ов 1.5 i 3j + 3 k The magnitude of the vector is obtained as 3 4 4.500 m D The unit vector along the direction of this vector is: B D B D 0.333 i 0.667 k 0.667j + the force can be written in vector form as: Since 999 FBD Fa D = F в о 666.333 k 1N 332.667 i 666.333 i + АВ х Fan about A Moment of FD j 0.00 3.00 0.00 -666.333 332,667 666.333 1998.999 i + 998.001 k] N m along B E Force F The force is directed from B to E whose coordinates are as follows: в ( 0.0 3, m E ( 1.5 0, 3 m Therefore, the position vectors are: 0 i+ 3 + ов н 0 0j + 1.5 i + 3 k OE= ВE 3 ОE - O B Hence, 1.5 3j+ of the vector is obtained as The 1.5 2 + -3 2 BE 3. 4.500 m

The unit vector along the direction of this vector is: ВE / B E 0.667 j 0.667 k 0.333 Since 1998 the force can be written in vector form as: = F FRE 1332.666 k1 N 665.334 i 1332.666 Moment of F ar about A АВх F BE j k 0.00 3.00 0.00 665.334 1332.666 1332.666 3997.998 i 0 j 1996.002 kN m + along C F Force F C to F The force is directed from whose coordinates are as follows: C ( F ( Therefore, the position vectors are: 0.0 6 m -3, 0, 2 m ос 0 i 6 j+ 0 k O F = 0 j+ -3 i + 2 k оF - -3 i O C Hence C F = 6j 2 k The magnitude of the vector is obtained as -3 2 -6 + 2 7.000 m The unit vector along the direction of this vector is: C F CF -0.429 i 0.857 j 0 286 k F E Since the force can be written in vector form as: 155 N F CF F c F 44.330 k 1 -66.495 132.835 + Moment otlong about A = А Сх C F j k 0.00 6.00 0.00 -66.495 44.330 -132.835 [265.980i + 398. 0 j + k] N m RESULTANT FORCE ABOUT A

and equating the components of i, j and k to zero. 332.667 i Summing up all the forces about A 666.333 k1 RA 666.333 + - 1332.666 j 1332.666 k 1 665.334 -66.495 i 132.835 44.330 + 931.506 i 2131.83 j 2043.329 k RESULTANT COUPLE ABOUT A and equating the components of i, j and k to zero. k Summing up all the moments about 1998.999 i 998,001 MA = 0 i + 1996.002 k1 3997.998 0.000 398.970 k] 265.980 i + 0.000 i + 931 506 0.000 2595.033 k