Solution: (a) (b) (N) F X =0: A X =0 (N) F Y =0: A Y + B Y (54)(9.81) 36(9.81)=0

Transcription

1 Prolem 5.6 The masses of the person and the diving oard are 54 kg and 36 kg, respectivel. ssume that the are in equilirium. (a) Draw the free-od diagram of the diving oard. () Determine the reactions at the supports and. W P W D 1.2 m 2.4 m 4.6 m Solution: (a) () (N) X =0: X =0 (N) Y =0: Y + Y (54)(9.81) 36(9.81)=0 M =0: 1.2 Y (2.4)(36)(9.81) (4.6)(54)(9.81)=0 W D W P Solving: X =0N Y = 1.85 kn Y =2.74 kn 1.2 m 2.4 m 4.6 m X 1.2 m 2.4 m 4.6 m Y Y W D W P Prolem 5.7 The ironing oard has supports at and that can e modeled as roller supports. (a) Draw the free-od diagram of the ironing oard. () Determine the reactions at and. 10 l 3 l 12 in 10 in 20 in Solution: The sstem is in equilirium. (a) The free-od diagram is shown. () The sums of the forces are: Sustitute into the force alance equation: =13 = l X =0, Y = =0. The sum of the moments aout is M =12 22(10) 42(3) = 0, 10 l 12 in 10 in 20 in 3 l from which = = in. 10 l 3 l

2 Prolem 5.17 With each of the devices shown ou can support a load R appling a force. The are called levers of the first, second, and third class. (a) The ratio R/ is called the mechanical advantage. Determine the mechanical advantage of each lever. () Determine the magnitude of the reaction at for each lever. (Epress our answer in terms of.) irst-class lever R R Second-class lever R Third-class lever Solution: ever of first kind. (a) The sum of the forces is Y = + R =0. R The sum of the moments aout is M = R =0, from which R = =1 () The reaction at is otained from the force alance equation: = R + =2 R (a) ever of second kind. The sum of forces is Y = R + =0. The sum of the moments aout is M = R +2 =0, R from which R = 2 =2 () The reaction at is otained from the force alance equation: = + R = +2 = ever of third kind. (a) The sum of forces is Y = R + =0. The sum of moments aout is: M = 2R + =0, from which: R = 2 = 1 2 () rom the force alance equation = + R = + 2 = 2, = 2

3 Prolem 5.27 The airplane s weight is W = 2400 l. Its rakes keep the rear wheels locked. The front (nose) wheel can turn freel, and so the ground eerts no horiontal force on it. The force T eerted the airplane s propeller is horiontal. (a) Draw the free-od diagram of the airplane. Determine the reaction eerted on the nose wheel and the total normal reaction on the rear wheels () when T =0, (c) when T = 250 l. 4 ft T 5 ft W 2 ft Solution: (a) The free od diagram is shown. () The sum of the forces: X = X =0 Y = Y W + Y =0. The sum of the moments aout is M = 5W +7 Y =0, from which Y = 5W = l 7 Sustitute from the force alance equation: Y = W Y = l (c) The sum of the forces: X = X =0, from which X = 250 l Y = Y W + Y =0. The sum of the moments aout : M = (250)(4) 5W +7 Y =0, from which Y = l. Sustitute into the force alance equation to otain: Y = l T 4 ft W 5 ft 2 ft 4 ft W Y 5 ft 2 ft Y X

4 Prolem 5.38 Determine the reactions at. 300 l 800 ft-l 5 ft 200 l 200 l 6 ft 3 ft Solution: The uilt-in support at is a two-force and couple reaction support. The sum of the forces for the sstem is X = X +200=0, from which X = 200 l Y = Y =0, from which Y = 100 l The sum of the moments aout : M = 6(300) + 5(200) M =0, from which M = 1600 ft l which is the couple at. 300 l 5 ft 800 ft-l 6 ft 3 ft 200 l 200 l Y M X 300 l 800 ft-l 5 ft 200 l 200 l 6 ft 3 ft

5 Prolem 5.50 Determine the reactions at the supports. 6 in 5 in 50 l 3 in 100 in-l 3 in 30 Solution: The reaction at is a two-force reaction. The reaction at is one-force, normal to the surface. The sum of the forces: X = X cos 60 50=0. Y = Y + sin 60 =0. 3 in 6 in 5 in 100 in l 50 l The sum of the moments aout is M = sin 60 6 cos 60 =0, 3 in 30 from which 100 = (11 sin 60 6 cos 60 =15.3l. ) Sustitute into the force equations to otain Y = sin 60 = 13.3 l and X = cos =57.7l X Y 11 in. 50 l 6 in Prolem 5.51 The weight W =2kN. Determine the tension in the cale and the reactions at. 30 W 0.6 m 0.6 m Solution: Equilirium Eqns: X =0: X + T cos 30 =0 Y =0: Y + T + T sin 30 W =0 + M =0: ( 0, 6)(W )+(0.6)(T sin 30 ) 30 Solving, we get +(1, 2)(T )=0 W 0.6 m 0.6 m X = 693 N, Y = 800 N, T = 800 N X Y T 30 T 0.6 m 0.6 m W = 2 kn = 2000 N

6 Prolem 5.60 The weight W 1 = 1000 l. Neglect the weight of the ar. The cale goes over a pulle at. Determine the weight W 2 and the reactions at the pin support. W W 2 Solution: The strateg is to resolve the tensions at the end of ar into - and -components, and then set the moment aout to ero. The angle etween the cale and the positive ais is 35. The tension vector in the cale is T 2 = W 2 (i cos( 35 )+j sin( 35 )). = W 2 (0.8192i j)(l). ssume a unit length for the ar. The angle etween the ar and the positive -ais is = 130. The position vector of the tip of the ar relative to is W r = i cos(130 )+j sin(130 ), = i j. The tension eerted W 1 is T 1 = 1000j. The sum of the moments aout is: M =(r T 1 )+(r T 2 )=r (T 1 + T 2 ) i j = W W M =( W )k =0, T 1 35 T 2 r 50 W 2 from which W 2 = l The sum of the forces: X =( X + W 2 (0.8192))i =0, X Y from which X = l Y =( Y W 2 (0.5736) 1000)j =0, from which Y = l

7 Prolem 5.66 onsider the eam in Prolem hoose supports at and so that it is not staticall indeterminate. Determine the reactions at the supports. Solution: One possiilit is shown: the pinned support at is replaced a roller support. The equilirium conditions are: X = X =0. 20 N-m The sum of moments aout is M = M +1.1 Y =0, from which Y = 20 = N. 1.1 The sum of forces in the vertical direction is Y = Y + Y =0, X Y 800 mm 300 mm 20 N-m 800 mm 300 mm Y from which Y = Y =18.18 N. Prolem 5.67 (a) Draw the free-od diagram of the eam and show that it is staticall indeterminate. (The eternal couple M 0 is known.) () an analsis of the eam s deflection, it is determined that the vertical reaction eerted the roller support is related to the couple M 0 =2M 0 /. What are the reactions at? M 0 Solution: (a) X : X = 0 (1) M 0 Y : Y + = 0 (2) + M : M M O + = 0 (3) Unknowns: M, X, Y,. 3 Eqns in 4 unknowns Statisticall indeterminate () Given =2M O / (4) Y M O We now have 4 eqns in 4 unknowns and can solve. Eqn (1) ields X =0 Eqn (2) and Eqn (4) ield M X Y = 2M O / Eqn (3) and Eqn (4) ield M = M O 2M O M = M O M was assumed counterclockwise M = M O clockwise X =0 Y = 2M O /

8 Prolem 5.77 The force eerted on the highwa sign wind and the sign s weight is = 800i 600j (N). Determine the reactions at the uilt-in support at O. Solution: The force acting on the sign is = X i + Y j + Z k = 800i 600j +0k N, and the position from O to the point on the sign where acts is r =0i +8j +8k m. TUON 8 m The force equations of equilirium for the sign are O X + X =0, O Y + Y =0, and O Z + Z =0. 8 m O Note that the weight of the sign is included in the components of. The moment equation, in vector form, is M = MO + r. Epanded, we get M = MOX i + M OY j + M OZ k =0. The corresponding scalar equations are M OX (8)( 600) = 0, M OY + (8)(800) = 0, 8 m and M OZ (8)(800) = 0. O Y Solving for the support reactions, we get M O O X = 800 N, O Y = 600 N, O Z =0, 8 m O Z O X M OX = 4800 N-m, M OY = 6400 N-m, and M OZ = 6400 N-m. Prolem 5.78 In Prolem 5.81, the force eerted on the sign wind and the sign s weight is = ±4.4v 2 i 600j (N), where v is the component of the wind s velocit perpendicular to the sign in meters per second (m/s). If ou want to design the sign to remain standing in hurricane winds with velocities v as high as 70 m/s, what reactions must the uilt-in support at O e designed to withstand? Solution: The magnitude of the wind component of the force on the sign is (4.4)(70) 2 N =21.56 kn. The force acting on the sign is = X i + Y j + Z k = ±21560i 600j +0k N, and the position from O to the point on the sign where acts is r = 0i +8j +8k m. The force equations of equilirium for the sign are O X + X =0, O Y + Y =0, and O Z + Z =0. Note that the weight of the sign is included in the components of. The moment equation, in vector form, is M = MO + r. Epanded, we get M = MOX i + M OY j + M OZ k ± =0. The corresponding scalar equations are M OX (8)( 600) = 0, M OY + (8)(±21560) = 0, and M OZ (8)(±21560) = 0. Solving for the support force reactions, we get O X = ±21560 N, O Y = 600 N, O Z =0, and for moments, M OX = 4800 N-m, M OY = ±172,500 N-m, and M OZ = ±172,500 N-m.

9 Prolem 5.87 The 158,000-kg airplane is at rest on the ground ( =0is ground level). The landing gear carriages are at,, and. The coordinates of the point G at which the weight of the plane acts are (3, 0.5, 5) m. What are the magnitudes of the normal reactions eerted on the landing gear the ground? 6 m 21 m 6 m G Solution: Y =(N + N R )+N W =0 MR = 3 mg +21N =0 21 m Solving, N = kn (1) (N + N R ) = kn (2) Y = N R + N + N W =0 6 m G (same equation as efore) 6 m + M O =0.5 W 6(N R )+6(N )=0(3) Solving (1), (2), and (3), we get N = kn N R = kn N = kn 3 m mg 21 m Side View R (N + N R ) Z N W 0.5 m ront View 6 6 N R N N

10 Prolem 5.93 In Prolem 5.95, the vertical force = 4 kn. The earing at will safel support a force of 2.5-kN magnitude and a couple of 0.5 kn-m magnitude. ased on these criteria, what is the allowale range of the distance? Solution: The solution to Pro produced the relations Y + =0 ( =4kN) 0.3 =0 M Z =0 X = Z = M X = M Y =0 Set the force at to its limit of 2.5 kn and solve for. In this case, M Z = 0.5 (kn-m) which is at the moment limit. The value for is = m We make Y unknown, unknown, and unknown ( =4kN, M Y = +0.5 (kn-m), and solve we get Y = 2.5 at = m However, 0.3 is the phsical limit of the device. Thus, m 0.3 m 0.3 m 0.2 m Prolem 5.94 The 1.1-m ar is supported a all and socket support at and the two smooth walls. The tension in the vertical cale D is 1 kn. (a) Draw the free-od diagram of the ar. () Determine the reactions at and. 400 mm Solution: (a) The all and socket cannot support a couple reaction, ut can support a three force reaction. The smooth surface supports oneforce normal to the surface. The cale supports one force parallel to the cale. () The strateg is to determine the moments aout, which will contain onl the unknown reaction at. This will require the position vectors of and D relative to, which in turn will require the unit vector parallel to the rod. The angle formed the ar with the horiontal is required to determine the coordinates of : ( ) 0.6 α = cos = The coordinates of the points are: (0.7, 0, 0.6), (0, 1.1 (sin 33.1 ), 0)=(0, 0.6, 0), from which the vector parallel to the ar is r = r r = 0.7i +0.6j 0.6k (m). The unit vector parallel to the ar is e = r = i j k. 1.1 The vector location of the point D relative to is 700 mm D 600 mm Epand and collect like terms: M =(0.6 Z )i (0.6 X 0.7 Z )j +( 0.6 X )k =0. rom which, Z = = kn, 0.6 X = = kn. 0.6 The reactions at are determined from the sums of the forces: X =( X + X )i =0, from which X = kn. Y =( Y 1)j =0, from which Y =1kN. Z =( Z + Z )k =0, from which Z = kn r D =( )e =0.7e = i j k. The reaction at is horiontal, with unknown -component and -components. The sum of the moments aout is M = r + r D D =0= X 0 Z =0 D 700 m 400 m 600 m T Z Y X

11 Prolem The plate is supported hinges at and and the cale E, and it is loaded the force at D. The edge of the plate to which the hinges are attached lies in the plane, and the aes of the hinges are parallel to the line through points and. The hinges do not eert couples on the plate. What is the tension in cale E? E 3 m 2i 6j (kn) D 1 m 2 m 20 2 m Solution: = + + D + T E =0 However, we just want tension in E. This quantit is the onl unknown in the moment equation aout the line. To get this, we need the unit vector along E. Point is at (2, 2 sin 20, 2 cos 20 ) Point E is at (0, 1, 3) e E = r E r E E 3 m 2i 6j (kn) D e E = 0.703i j k We also need the unit vector e. (0, 0, 0), (0, 2 sin 20, 2 cos 20 ) e =0i 0.342j k 1 m 20 2 m 2 m The moment of D aout (a point on ) is M D = r D D1 =(2i) (2i 6j) M D = 12k The moment of T E aout (another point on line E)is M TE = r T E e E =2i T E e E, where e E is given aove. The moment of D aout line is M D = M D e M D = N-m E 1 m Y 20 Z 3 m X 2 m D = 2i 6j Y X D Z T E 2 m The moment of T E aout line is M E M E = T E (2i e E ) e = T E ( 0.788j k) e M E =1.382T E The sum of the moments aout line is ero. Hence M D + M E = T E =0 T E =8.15 kn

12 Prolem The earings at,, and do not eert couples on the ar and do not eert forces in the direction of the ais of the ar. Determine the reactions at the earings due to the two forces on the ar. 300 mm 200 i (N) 180 mm 150 mm 100 k (N) 150 mm Solution: The strateg is to take the moments aout and solve the resulting simultaneous equations. The position vectors of the earings relative to are: 300 mm 200 i (N) r = 0.15i +0.15j, r = 0.15i +0.33j +0.3k. Denote the lower force suscript 1, and the upper suscript 2: 180 mm r 1 = 0.15i, r 2 = 0.15i +0.33j. The sum of the moments aout is: M = r r + r r =0 M = X 0 Z X Y 0 =0 M =(0.15 Z 0.3 Y )i +( Z +0.3 X )j 100 k (N) Y X Z 150 mm 200 N X Y 150 mm +( 0.15 X Y 0.33 X )k =0. This results in three equations in four unknowns; an additional equation is provided the sum of the forces in the -direction (which cannot have a reaction term due to ) X =( X + X + 200)i = N Z The four equations in four unknowns: 0 X Z +0 X 0.3 Z =0 0 X Z +0.3 X +0 Y = X +0 Z 0.33 X 0.15 Y =66 X +0 Z + X +0 Z = 200. (The HP-28S hand held calculator was used to solve these equations.) The solution: X = 750 N, Z = 1800 N, X = 950 N, Y = 900 N. The reactions at are determined the sums of forces: Y =( Y + Y )j =0, from which Y = Y = 900 N Z =( Z + Z + 100)k =0, from which Z = 1900 N

13 Prolem rectangular plate is sujected to two forces and (ig. a). In ig., the two forces are resolved into components. writing equilirium equations in terms of the components,,, and, show that the two forces and are equal in magnitude, opposite in direction, and directed along the line etween their points of application. h (a) h () Solution: The sum of forces: X = X + X =0, from which X = X Y = Y + Y =0, h ig a from which Y =. These last two equations show that and are equal and opposite in direction, (if the components are equal and opposite, the vectors are equal and opposite). To show that the two vectors act along the line connecting the two points, determine the angle of the vectors relative to the positive -ais. The sum of the moments aout is ig M = (h) =0, from which the angle of direction of is ( ) ( ) tan 1 Y h = tan 1 = α. X or (180 + α ). Similarl, sustituting : ( ) ( ) tan 1 Y h = tan 1 = α, X or (180 + α ). ut ) α = tan 1 ( h descries direction of the line from to. The two vectors are opposite in direction, therefore the angles of direction of the vectors is one of two possiilities: is directed along the line from to, and is directed along the same line, oppositel to.

14 Prolem To determine the location of the point where the weight of a car acts (the center of mass), an engineer places the car on scales and measures the normal reactions at the wheels for two values of α, otaining the following results: α (kn) (kn) 10 W What are the distances and h? α 2.7 m h Solution: The position vectors of the cm and the point are r M =(2.7 )i + hj, r =2.7i. The angle etween the weight and the positive -ais is β = 270 α. The weight vector at each of the two angles is W 10 = W (i cos j sin 260 ) W 10 = W ( i j) W 20 = W (i cos j sin 250 ) or W 20 = W ( i j) The weight W is found from the sum of forces: Y = Y + Y + W sin β =0, from which W β = Y + Y. sin β Taking the values from the tale of measurements: W 10 = sin 260 = kn, These two simultaneous equations in two unknowns were solved using the HP- 28S hand held calculator. =1.80 m, h =0.50 m α W 2.7 m h [check :W 20 = sin 250 = kn check] The moments aout are M = r M W + r =0. Taking the values at the two angles: M 10 = 2.7 h =0 = h = 0 M 20 = 2.7 h = h = 0