5.3 Rigid Bodies in Three-Dimensional Force Systems

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1 5.3 Rigid odies in Three-imensional Force Sstems

2 5.3 Rigid odies in Three-imensional Force Sstems Eample 1, page 1 of 5 1. For the rigid frame shown, determine the reactions at the knife-edge supports,,. Neglect the weight of the frame. 80 N 150 mm 200 mm E 150 mm 24 N 400 mm 600 mm 300 mm

3 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 1, page 2 of 5 1 Free-bod diagram of frame. 80 N F F E 24 N F 2 Since all forces are vertical and dimensions are given in coordinate directions, a scalar rather than a vector approach is probabl best. 3 Sum forces F = 0: F + F + F N = 0 (1)

4 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 1, page 3 of 5 80 N 4 To calculate moments acting about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. 150 mm 200 mm F F E 150 mm 24 N F 400 mm 600 mm 300 mm 6 Sum moments about the ais. 150 mm 5 View from the positive ais 150 mm M = 0: Solving gives F (150 mm mm) + 24 N (200 mm) + 80 N (150 mm) = 0 E 80 N,, F = 56.0 N F 24 N F 200 mm F

5 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 1, page 4 of 5 80 N 7 To calculate moments about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. 150 mm 200 mm F F E 150 mm 24 N F 400 mm 600 mm 300 mm 9 Sum moments about the ais. 80 N 8 View from the positive ais. M = 0: 80 (400 mm) + 56 N (400 mm) + F (400 mm mm) 24 N (400 mm mm mm) = 0, E Solving gives F F = 56.0 N F 24 N F = 40.8 N 400 mm 600 mm 300 mm

6 5.3 Rigid odies in Three-imensional Force Sstems Eample 1, page 5 of 5 10 Using F = 56.0 N and F = 40.8 N in Eq. 1 gives F + F + F 80 N 60 N = 0 (Eq. 1 repeated) 56.0 N 40.8 N Solving gives F = 7.2 N

7 5.3 Rigid odies in Three-imensional Force Sstems Eample 2, page 1 of kg block rests on top of a triangular plate of negligible weight. The plate is supported b vertical wires, O, and E. etermine the and coordinates of the 10-kg block if the tensions in wires and E both equal 20 N. lso determine the tension in wire O. 400 mm 300 mm O E 10 kg

8 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 2, page 2 of 4 1 Free-bod diagram of plate and 10-kg block. 400 mm T O 20 N 300 mm O E 20 N 10 kg Weight = (10 kg)(9.81 m/s2) = N 3 2 Since all forces are parallel to a single coordinate () ais and dimensions are given parallel to coordinate directions, a scalar rather vector approach is probabl best. Sum forces F = 0: T O + 20 N + 20 N N 0 Solving gives T O = 58.1 N ns

9 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 2, page 3 of 4 4 To calculate and, we need to use moment equations of equilibrium. 400 mm 300 mm T O O N 20 N E 20 N 5 To calculate moments about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. 20 N 6 View from the positive ais 300 mm N 7 Sum moments about the positive ais. M = 0: (98.10 N)() (20 N)(300 mm) = 0, E, O Solving gives = 61.2 mm

10 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 2, page 4 of mm T O 20 N 300 mm O N E 20 N 8 To calculate moments about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane N T O View from the positive ais N 20 N 10 Sum moments about the ais. M = 0: 20 N (400 mm) (98.10 N) = 0 Solving gives,, O E = 81.6 mm 400 mm

11 5.3 Rigid odies in Three-imensional Force Sstems Eample 3, page 1 of 6 3. To hoist the 180-lb load, a vertical force P is applied to the crank of the windlass shown. The bearing at eerts forces normal and parallel to the shaft; the bearing at eerts onl forces normal to the shaft no aial force. etermine the magnitude of P and the forces eerted b the bearings when the crank is in the position shown. 12 in. 24 in. 10 in. earing earing 60 Radius = 8 in. 180 lb P P 18 in. View as seen from the right end.

12 Rigid odies in Three-imensional Force Sstems Eample 3, page 2 of 6 1 Free-bod diagram of windlass 12 in. 24 in. 10 in. 3 Sum forces. 180 lb 2 No component of force is shown because the bearing transmits no aial force P F = 0: = 0 (1) F = 0: lb + P 0 2) F = 0: + = 0 (3)

13 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 3, page 3 of 6 12 in. 24 in. 10 in. 4 To calculate moments acting about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. 180 lb P 5 View from the positive ais (18 in.) cos 60 = 9 in. Radius = 8 in.,, 6 M = 0: (180 lb)(8 in.) P(9 in.) = 0 60 Solving gives P = 160 lb P 180 lb 18 in.

14 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 3, page 4 of 6 7 To calculate the moments acting about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. 12 in. 24 in. 10 in. 180 lb 36 in. P 9 Sum moments about the ais., M = 0: (36 in.) = 0 8 View from the positive ais So = 0

15 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 3, page 5 of 6 12 in. 24 in. 10 in. 180 lb 10 To calculate the moments about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. P, 11 View from the positive ais 12 Sum moments about the ais. 180 lb M = 0: (180 lb)(12 in.) + (12 in in.) (160 lb)(12 in in in.) = 0 P = 160 lb Solving gives 12 in. 24 in. 10 in. = lb

16 5.3 Rigid odies in Three-imensional Force Sstems Eample 3, page 6 of 6 13 Substituting P = 160 lb, = lb, and = 0 in Eqs. 1, 2, and 3 and solving gives = 0 = lb = 0

17 5.3 Rigid odies in Three-imensional Force Sstems Eample 4, page 1 of 6 4. The bo is supported b short links at the corners. etermine the forces in the links. 30 G H 20 lb 5 ft 42 lb O E 30 6 ft 6 ft 4 ft

18 Rigid odies in Three-imensional Force Sstems Eample 4, page 2 of 6 1 Free-bod diagram of bo F G 30 G H 20 lb 5 ft F 42 lb O E 30 F O F E 6 ft 2 Sum forces. F = 0: 20 lb F cos 30 F G cos 30 = 0 (1) F = 0: F sin 30 + F G sin 30 F F O F E 0 (2) F F 6 ft F = 0: F + 42 lb = 0 (3) 4 ft

19 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 4, page 3 of 6 3 To calculate the moments about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. F G 30 G H F 42 lb O 20 lb E 5 ft 4 F sin lb View from the positive ais H,G 30 F O F E 5 ft, 6 ft, 12 ft,e,o F F 6 ft F 5 Sum moments about the ais. 4 ft M = 0: F (12 ft) F sin 30 (12 ft) + 42 lb (5 ft) = 0 (4)

20 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 4, page 4 of 6 F G 30 G 6 To calculate the moments about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. H,G,O 4 ft H,E 6 ft 20 lb 5 ft 20 lb F lb F O O E F E F cos 30, 42 lb, 6 ft F 6 ft 7 View from the positive ais F F 4 ft 6 ft 8 Sum moments about the ais. M = 0: F cos 30 (6 ft + 6 ft) F (4 ft) 20 lb (6 ft) 42 lb (4 ft) = 0 (5)

21 + 5.3 Rigid odies in Three-imensional Force Sstems Eample 4, page 5 of 6 9 To calculate moments about the ais, consider the view seen b an observer located on the positive ais and looking back at the plane. F cos 30 F G cos 30,G,H 20 lb F G 30 G H 5 ft 20 lb 5 ft,,o,e F 42 lb O E 4 ft F E 30 F O F E 10 View from the positive ais F F 6 ft 6 ft 11 Sum the moments about the ais. M = 0: F G cos 30 (5 ft) + F cos 30 (5 ft) 20 lb (5 ft) F E (4 ft) 0 (6) 4 ft

22 5.3 Rigid odies in Three-imensional Force Sstems Eample 4, page 6 of 6 12 Solving the si equilibrium equations 20 F cos 30 F G cos 30 = 0 (1) F sin 30 + F G sin 30 F F O F E = 0 2 F + 42 = 0 (3) (42)(5) (12 F ) sin F = 0 (4) (12 F ) cos 30 4F + (6)(20) (42)(4) = 0 (5) (5 F ) cos 30 + (5 F G ) cos 30 (20)(5) 4F E = 0 (6) gives F = 11.7 lb F = 42.0 lb F = 11.5 lb F E = 0 F G = 11.5 lb F O = 23.3 lb ns

23 5.3 Rigid odies in Three-imensional Force Sstems Eample 5, page 1 of 6 5. force F = {20i 40j 30k} N acts at the midpoint of rod. End of the rod rests on a smooth horiontal plate and is attached to cord. End is attached to a ball-and-socket support. etermine the force in the cord and the reactions at and. F = {20i 40j + 30k} N 3 m ord 4 m 5 m

24 5.3 Rigid odies in Three-imensional Force Sstems Eample 5, page 2 of 6 k 1 Free-bod diagram of rod i 2,, and are the forces from the ball-and-socket joint at. No moment is transmitted at a ball-and-socket joint. 3 m j F = {20i 40j + 30k} N O 4 m Tk 3 Tension in cord 5 m j 4 Normal force from plate 5 Sum forces. F = 0: i + j + k + j + {20i 40j + 30k}N Tk = 0 ollecting the coefficients of i, j, and k gives ( + 20)i + ( + 40)j + ( + 30 T)k = 0 6 Equating coefficients to ero gives F = 0: + 20 = 0 (1) F = 0: + 40 = 0 (2) F = 0: + 30 T = 0 (3)

25 5.3 Rigid odies in Three-imensional Force Sstems Eample 5, page 3 of 6 7 Net, calculate moments. To do so, first introduce position vectors from to and from to. k i 9 r = position vector from to midpoint of rod = 2 r 3 m O j F = {20i 40j + 30k} N 8 r = {5i 3j + 4k} m = {5i 3j + 4k} 2 4 m Tk = {2.5i 1.5j + 2k} m 5 m j

26 5.3 Rigid odies in Three-imensional Force Sstems Eample 5, page 4 of 6 10 Sum moments about point. M = 0: r F + r {j k} = 0 or, {2.5i 1.5j + 2k} {20i 40j 30k} + { i 3j+ 4k} {j k} = 0 Evaluating the cross products gives i j k i j k 0 = T = i j + k i j + k T 0 T 0 = i[( 1.5)(30) ( 40)(2)] j[2.5(30) 20(2)] + k[2.5( 40) 20( 1.5)] + i[( 3)( T) (4)] j[5( T) 0(4)] + k[5() 0)( 3)]

27 5.3 Rigid odies in Three-imensional Force Sstems Eample 5, page 5 of 6 11 arring out the multiplications and collecting coefficients of i, j, and k gives 12 Substituting T = 14 N and = 8 N and solving Eqs. 1, 2, and 3 gives 0 = (35 + 3T 4)i + ( T)j + ( )k Equating each coefficient to ero gives M = 0: T 4 = 0 (4) M = 0: T = 0 (5) M = 0: = 0 (6) Solving Eq. 5 gives = 20 N = 26 N = 23 N T = 7 N and solving either Eq. 4 or Eq. 6 (one of the equations are redundant) gives = 14 N Eqs. 1, 2, and 3 are F = 0: + 20 = 0 (Eq. 1 repeated) F = 0: + 40 = 0 (Eq. 2 repeated) F = 0: + 30 T = 0 (Eq. 3 repeated)

28 5.3 Rigid odies in Three-imensional Force Sstems Eample 5, page 6 of 6 ' k ' i j F O Tk j ' 13 Wh is one of the moment equations, M = 0 and M = 0, redundant? ecause the lines of action of all forces pass through the ais of the rod. Thus if, instead of,, and aes, we had chosen ', ', and ' aes as shown above, then since all forces intersect the ' ais, the moment equation M' = 0 would reduce to 0 = 0. That is, one of the moment equations is not an independent equation, and there are onl two non-trivial moment equations: M ' = 0 and M' = 0. General conclusion: In three dimensions, there are at most si independent equations of equilibrium.

29 5.3 Rigid odies in Three-imensional Force Sstems Eample 6, page 1 of 6 6. The boom is supported b a ball-and-socket at and gu wires F and E. etermine the components of the reactions at and the forces in the gu wires. 1 m 4 m 2 m 3 m E F 2 m 2 m 50 kg 3 m

30 5.3 Rigid odies in Three-imensional Force Sstems Eample 6, page 2 of 6 1 Free-bod diagram k F F F E i E j 2,, and components of force at a ball-and-socket joint F able force supporting 50-kg block = 50 kg 9.81 m/s 2 = N 3 Sum forces. F = 0: F F + F E + i + j + k {490.5j} N = 0 (1) 4 Net we have to epress F F and F E in rectangular components.

31 5.3 Rigid odies in Three-imensional Force Sstems Eample 6, page 3 of 6 2 m r F F F 4 m F E 1 m r E 3 m 5 Introduce position vectors in the same direction as the forces. r F = ( 4 m 0)i + (2 m 0)j + (0 2 m 2 m)k = { 4i + 2j 4k}m r E = (1 m 0)i + (3 m 0)j + (0 2 m)k F E = {i + 3j 2k}m 50 kg 3 m 2 m 2 m 6 Force = (magnitude)(unit vector): F F r F r F = (F F )( ) 4i + 2j 4k = F F ( ) ( 4) 2 + (2) 2 + ( 4) 2 = F F { i j k} (2)

32 5.3 Rigid odies in Three-imensional Force Sstems Eample 6, page 4 of 6 F E r E r E = (F E )( ) = F E ( i + 3j 2k ) (1) 2 + (3) 2 + ( 2) 2 = F E {0.2673i j k) (3) 7 Substitute Eqs. 2 and 3 into the equilibrium equation, Eq. 1: F F + F E + i + j + k (490.5)j = 0 (Eq. 1 repeated) F E {0.2673i j k} F F { i j k} ollecting coefficients of i, then of j, and then of k gives ( F F F E + )i + (0.3333F F F E )j + ( F F F E + )k = 0 Equating each coefficient to ero gives F = 0: F F F E + = 0 (4) F = 0: F F F E = 0 (5) F = 0: F F F E + = 0 (6)

33 5.3 Rigid odies in Three-imensional Force Sstems Eample 6, page 5 of 6 F F F r F r E E F E 8 To calculate moments, first introduce position vectors r E = {2k} m r F = (2m + 2m)k = {4k} m r = (2m + 2m + 3m)k = {7k} m r 2 m 2 m 9 Sum moments. or, M = 0: r E F E + r F F F + r { 490.5j} = 0 3 m {2k} F E {0.2673i j k} + {4k} F F { i j k} + {7k} { 490.5j} = 0 Epanding the cross products gives 2F E (0.2673)k i + 2F E (0.8018)k j + 2F E ( )k k + 4F F ( )k i = j = i = 0 = j + 4F F (0.3333)k j + 4F F ( )k k + 7( k j = 0 = i = 0 = i

34 5.3 Rigid odies in Three-imensional Force Sstems Eample 6, page 6 of 6 10 arring out the multiplications and collecting coefficients of i and then j gives ( F E F F )i + (0.5334F E F F )j = 0 Equating each coefficient to ero gives M = 0:.6036F E F F = 0 (7) M = 0: F E F F = 0 (8) Solving Eqs. 7 and 8 simultaneousl gives F E = N = kn F F = N = kn Using these results in Eqs. 4, 5, and 6 gives = N = kn = N = kn = N = kn

35 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 1 of 8 7. The uniform 15-kg lid of the bo is supported b a stick and short hinges at and. ssuming that hinge transmits no aial force and that the line of action of the force from the stick coincides with the long ais of the stick, determine the reactions at the hinges and the force transmitted b the stick at. 250 mm 250 mm 400 mm mm 30 mm

36 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 2 of 8 1 Free-bod diagram of lid 2 Weight = (15 kg)(9.81 m/s 2 ) = N 3 F = force from stick acting on lid at point. 400 mm = mm mm k F j i E (center of lid) r 30 mm 440 mm 30 mm k j 400 mm 400 mm 4 = 0 because we are told that hinge transmits no aial force 5 6 Sum forces. F = 0: i + j + k + j + k + F {147.15j} N = 0 (1) The force F must be epressed in rectangular components. We know that, in terms of the position vector r : F = F r r

37 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 3 of mm r 400 sin 50 mm = mm 400 mm 50,, 500 mm View of plane as seen from positive ais 400 cos 50 mm = mm 400 mm 7 The force F must be epressed in rectangular components. We know that, in terms of the position vector r : Here, r F = F (2) r r = coordinates of coordinates of = {( )i + ( )j + ( )k} m View of plane as seen from positive ais 8 Thus, Eq. 2 gives F = F = F r r { 250i j 142.9k} ( 250) 2 + (306.4) 2 + ( 142.9) 2 = F { i j k} (3)

38 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 4 of 8 9 This epression for F can now be substituted into Eq. 1: i + j + k + j + k + F {147.15j} = 0 (Eq. 1 repeated) ollecting coefficients of i, then j, and then k gives F { i j k ( F )i + ( F )j + ( F )k = 0 Equating each coefficient to ero gives F = 0: F = 0 (4) F = 0: F = 0 (5) F = 0: F = 0 (6)

39 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 5 of 8 10 Net, moment equations must be written. If we choose point as the point about which to sum moments, then the resulting equations will be simplified, since,, and will not appear. Thus we need the position vectors from to the lines of action of the forces. 440 mm 440 mm 2 r E = 220 mm E N r j k 400 mm 11 From the figures, enter of lid 200 sin 50 mm = mm E mm 2 = 200 mm View of the plane as seen from the positive ais. 200 cos 50 mm = mm,, r E = {220i j 128.6k} mm (7) r = {440i} mm (8)

40 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 6 of 8 F E j 12 Line of action of force F r k 400 mm 440 mm 30 mm 13 We can use position vector r from to (to calculate the moment of the force F ) because the line of action of F passes through. r = {( )i +400k} mm (9)

41 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 7 of 8 14 Sum moments about. M = r E { j}N + r { j + k} + r F Substituting from Eqs. 8, 9, and 10 for r E, r, and r and from Eq. 3 for F gives M = {220i 153.2j 128.6k} { j} + {440i} { j k} + {470i 400k} F { i j 0 k} Epanding the cross products gives M = 220( )i j ( )j j ( )k j + 440( )i j + 440()i k = k = 0 = i = k = j + 470( F )i i + 470(0.7287F i j + 470( F )i k = 0 = k = j + 400( F )k i + 400(0.7287F )k j + 400( F )k k = j = i = 0

42 5.3 Rigid odies in Three-imensional Force Sstems Eample 7, page 8 of 8 15 arring out the multiplications and then collecting coefficients of i, j, and k gives M = ( F )i + ( F )j + ( F )k Setting coefficients of i, j, and k to ero gives F = 0 (10) F = 0 (11) F = 0 (12) Solving Eqs. 10, 11, and 12 simultaneousl gives F = N = N = N Substituting these results in Eqs. 4, 5, and 6, and then solving gives = 38.6 N = 76.8 N = 33.6 N

43 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 1 of 8 8. The uniform bar weighs 20-lb, is 4-ft long, and is supported b two cords and b smooth surfaces at and. etermine the forces in the cords and the reactions at and. ord 4 ft O ord

44 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 2 of 8 1 Free-bod diagram of N k (normal force onl, "smooth" surface so no friction force no force parallel to the wall) T i (tension in cord) O M (midpoint of ) { 20j} lb (weight) T Tension in cord N j (normal force onl, because surface is smooth) 2 Sum forces. F = 0: N k + T i + T + N j + { 20j} lb = 0 (1)

45 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 3 of 8 3 To find the rectangular components of the force T, introduce a position vector r O from to O. ord Magnitude r O = 4 ft cos 60 4 ft = 2 ft O r O component of r O = (2 ft) cos 30 = ft ord T 7 T = (T )(unit vector along O) r O = T r O 5 component of r O = (2 ft) sin 30 = 1 ft = T i k 2 = T i 0.5T k 6 Since the and components of r O are now known, r O can be epressed as r O = { i k} ft. (2)

46 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 4 of 8 8 The epression that has just been found for T in terms of rectangular components can now be substituted into the sum of forces equation, Eq. 1: N k + T i + T + N j 20j = 0 (Eq. 1 repeated) T i 0.5T k ollecting coefficients of i, then j, and then k gives (T T )i + ( N 20)j + (N 0.5T )k = 0 Each coefficient must equal ero so T T = 0 (3) N 20 = 0 (4) N 0.5T = 0 (5) Three equations, four unknowns. n additional equation is needed and will be obtained b writing a moment equation.

47 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 5 of To write the moment equation, first introduce position vectors from O to, from O to, and from O to the line of action of the weight. r O = (2ft + 2ft) sin 60 j 11 The position vector from O to, r O, can be found b noting that it is just the negative of the position vector from to O, r O, and we have alread found r O ( Eq. 2): r O = r O Eq. 2 = {3.4641j} ft (6) = { i k} = {1.7321i + k}ft (7) r O 2 ft M 12 MM' is the line of action of the weight of the rod. similar triangles MM' and O, point M' must lie at the midpoint of O (because M lies at the midpoint of the rod). Thus 1 r OM' = 2 r O O r OM' 2 ft 60 M' = 1 2 {1.7321i + k} = {0.8660i + 0.5k} ft (8) r O

48 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 6 of 8 N k T i r O M O r OM' M' T { 20j} lb r O N j 13 Sum moments about the origin O. = 0 because r O and T are parallel M O = 0: r O N k + T i) + r OM' { 20j} + (r O T ) + r O N j = 0 {3.4641j} N k + T i) + {0.8660i + 0.5k} { 20j} + {1.7321i + k} N j = 0

49 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 7 of 8 14 arring out the cross products gives N j k T j i ( 20)i j + 0.5( 20)k j N i j + N k j = 0 = i = k = k = i = k = i arring out the multiplications and collecting coefficients of i, then j, and then k gives (3.4641N + 10 N )i + ( T N )k = 0 Equating coefficients of i and then of k to ero gives N + 10 N = 0 (9) T N = 0 (10) These equations must be solved simultaneousl with Eqs. 3, 4, and 5: T T = 0 (Eq. 3 repeated) N 20 = 0 (Eq. 4 repeated) N 0.5T = 0 (Eq. 5s repeated)

50 5.3 Rigid odies in Three-imensional Force Sstems Eample 8, page 8 of 8 15 Eqs. 9, 10, 3, 4, and 5 constitute five equations in four unknowns. One equation must be redundant. Solving these equations gives N = 20 lb N = 2.89 lb T = 5 lb T = 5.77 lb

PROBLEMS. m s TAC. m = 60 kg/m, determine the tension in the two supporting cables and the reaction at D.

PROBLEMS. m s TAC. m = 60 kg/m, determine the tension in the two supporting cables and the reaction at D. 1. he uniform I-beam has a mass of 60 kg per meter of its length. Determine the tension in the two supporting cables and the reaction at D. (3/62) A( 500) m (5 23) m m = 60 kg/m determine the tension in