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Statics of Rigid Bodies: Equilibrium and Equivalent Force Systems, Exercises of Statics

Key CollegeStatics

Statics of Rigid Bodies | Equilibrium | Equivalent Force of Systems

Typology: Exercises

2017/2018

Available from 12/11/2021

Amsterdamn
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STATICS OF RIGID BODIES | EQUILIBRIUM
Sample Problem 1. A fixed crane has a mass of 1000 kg and is used to lift a 2400-kg crate. It is
held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G.
Determine the components of the reactions at A and B.
Solution.
A free-body diagram of the crane is drawn. By multiplying the masses of the crane and of the
crate by g = 9.81 m/s2, we obtain the corresponding weights, that is, 9810 N or 9.81 kN, and
23,500 N or 23.5 kN. The reaction at pin A is a force of unknown direction; it is represented by
its components Ax and Ay. The reaction at the rocker B is perpendicular to the rocker surface;
thus, it is horizontal. We assume that Ax, Ay, and B act in the directions shown.
Determination of B. We express that the sum of the moments of all external forces about point A
is zero. The equation obtained will contain neither Ax nor Ay, since the moments of Ax and Ay
about A are zero. Multiplying the magnitude of each force by its perpendicular distance from A,
we write
∑MA =0: B(1.5m)-(9.81kN)(2m)-(23.5kN)(6m)=0 B = 107.1 kN
Since the result is positive, the reaction is directed as assumed.
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STATICS OF RIGID BODIES | EQUILIBRIUM

Sample Problem 1. A fixed crane has a mass of 1000 kg and is used to lift a 2400-kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. Solution. A free-body diagram of the crane is drawn. By multiplying the masses of the crane and of the crate by g = 9.81 m/s 2 , we obtain the corresponding weights, that is, 9810 N or 9.81 kN, and 23,500 N or 23.5 kN. The reaction at pin A is a force of unknown direction; it is represented by its components Ax and Ay. The reaction at the rocker B is perpendicular to the rocker surface; thus, it is horizontal. We assume that Ax, Ay, and B act in the directions shown. Determination of B. We express that the sum of the moments of all external forces about point A is zero. The equation obtained will contain neither Ax nor Ay, since the moments of Ax and Ay about A are zero. Multiplying the magnitude of each force by its perpendicular distance from A, we write ∑MA =0: B(1.5m)-(9.81kN)(2m)-(23.5kN)(6m)=0 B = 107.1 kN Since the result is positive, the reaction is directed as assumed.

Determination of Ax. The magnitude of Ax is determined by expressing that the sum of the horizontal components of all external forces is zero. ∑Fx=0: Ax+B=0=Ax+107.1kN=0 Ax = -107.1 kN Since the result is negative, the sense of Ax is opposite to that assumed originally. Determination of Ay. The sum of the vertical components must also equal zero. ∑Fy=0: Ay-9.81kN-23.5kN=0 Ay = 33.3 kN Adding vectorially the components Ax and Ay, the reaction at A is 112.2 kN at 17.3° (W of N). Check. The values obtained for the reactions can be checked by recalling that the sum of the moments of all of the external forces about any point must be zero. For example, considering point B, we write ∑MB = -(9.81 kN)(2 m) - (23.5 kN)(6 m) + (107.1 kN)(1.5 m) = 0 Sample Problem 2. Three loads are applied to a beam as shown. The beam is supported by a roller at A and by a pin at B. Neglecting the weight of the beam, determine the reactions at A and B when P = 15 kips. Solution. A free-body diagram of the beam is drawn. The reaction at A is vertical and is denoted by A. The reaction at B is represented by components Bx and By. Each component is assumed to act in the direction shown. Equilibrium Equations. We write the following three equilibrium equations and solve for the reactions indicated: ∑Fx = 0: Bx = 0 ∑MA = 0: -(15 kips)(3 ft) + By(9 ft) - (6 kips)(11 ft) - (6 kips)(13 ft) = 0 By = 21.0 kips

Exercise Problems:

  1. A load of lumber of weight W = 25 kN is being raised by a mobile crane. The weight of boom ABC and the combined weight of the truck and driver are as shown. Determine the reaction at each of the two (a) front wheels H, (b) rear wheels K. (Ans. H = 34.04 kN each, K = 4.96 kN each)
  2. The frame shown supports part of the roof of a small building. Knowing that the tension in the cable is 150 kN, determine the reaction at the fixed end E.
  3. A loading car is at rest on a track forming an angle of 25° with the vertical. The gross weight of the car and its load is 25 kN, and it acts at a point 750 mm from the track, halfway between the two axles. The car is held by a cable attached 600 mm from the track. Determine the tension in the cable and the reaction at each pair of wheels. (Ans. T = 22.66 kN, R 1 = 2.56 kN, R 2 = 8 kN)
  1. A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. Ans. (a) 1. kN, (b) 166.3 N
  1. Collar A is connected as shown to a 200-N load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 90 mm, (b) x = 300 mm. (Hint: The tension in the rope is the same on each side of a simple pulley) Ans. (a) 43.9 N, (b) 120 N
  2. Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. (Ans. (a) 2.50 kN, (b) 2.72 kN)
  1. Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C. (Ans. DE = 600 N , C = 1252.84 N 69.83° N of E)
  2. A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA =8kNandFB =16kN,determinethe magnitudes of the other two forces. (Ans. FC = 6.40 kN, FD = 4.80 kN)

T 1 =3700lb,T 2 =2600lb Trigonometric Solution. The triangle rule can be used. We note that the triangle shown represents half of the parallelogram shown above. Using the law of sines, we write T 1 / sin 45° = T 2 / sin 30° = 5000 lb / sin 105° Simplifying the equation, we obtain T 1 =3660lbT 2 =2590lb b. Value of a for Minimum T 2. To determine the value of a for which the tension in rope 2 is minimum, the triangle rule is again used. In the sketch shown, line 1-1’ is the known direction of T 1. Several possible directions of T 2 are shown by the lines 2-2’. We note that the minimum value of T 2 occurs when T 1 and T 2 are perpendicular. The minimum value of T 2 is T 2 = (5000 lb) sin 30° = 2500 lb Corresponding values of T 1 and α are T 1 = (5000 lb) cos 30° = 4330 lb α = 90° - 30° α = 60°

Exercise problem:

  1. A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. (Ans. P = 392#, R = 346#)
  2. Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. (Ans. R= 1400 N, 48° N of E)
  3. Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. (Ans. R = 20 kN, 20° S of W)

Sample Problem. Four tugboats are bringing an ocean liner to its pier. Each tugboat exerts a 5000# force in the direction shown. Determine (a) the equivalent force couple system at the foremast O, (b) the point on the hull where a single, more powerful tugboat should push to produce the same effect as the original four tugboats. Solution. a. Force-Couple System at O. Each of the given forces is resolved into components. The force- couple system at O equivalent to their resultant force R and a couple, the moment of which is equal to MO as follows: R 2 =[5000x(cos60°+3/5+0+cos45°)] 2 +[5000x(-sin60°-4/5-1+sin45°)] 2 R 2 = (9035.53) 2

  • (- 9794.59)^2 , R = 13325.72 kips MO =5000x(50’xcos60°+70’x3/5+0-70xcos45°)+5000x(-90’xsin60°+100’x4/5 + 400’ x 1 -300’ x sin 45°) = 1037.14 kip-ft b. Single Tugboat. The force exerted by a single tugboat must be equal to R, and its point of application A must be such that the moment of R about O is equal to MO: MO =Rx·70’+Ry·x 1037.14 kip-ft = 9035.53 · 70’ + 9794.59 x x = 41.31 ft

Exercise Problems:

  1. In figure, assuming clockwise moments as positive, compute the moment of a force F = 450 lb and of force P = 361 lb about points A, B, C, and D. (Ans. F: MA = 1350 lb-ft CCW, MB = 2160 lb-ft CW, Mc = 1350 lb-ft CW, MD = 810 lb-ft CW; P: MA = 1.53 x 10
  • lb- ft CW or 0, MB = 300 lb-ft CCW, Mc = 1201.48 lb-ft CCW, MD = 1201. lb-ft CW)
  1. Locate the amount and position of the resultant of the loads acting on the Fink truss shown in figure. (Ans. R = 3400 lb downward, d = 12. 06 ft to the right of A)
  2. Determine the resultant of the four parallel forces acting on the rocker arm. (Ans. R = 50 lb downward, d = 4 ft to the right of O)
  1. A worker tries to move a rock by applying a 360-N force to a steel bar as shown. If two workers attempt to move the same rock by applying a force at A and a parallel force at C, determine these two forces so that they will be equivalent to the single 360-N force shown in the figure. (Ans. A = 168 N 50° S of W, C = 192 N 50° S of W)
  2. Compute the resultant of the three forces shown in figure. Locate its intersection with X and Y axes. (Ans. R = 957.74 lb, 32.17° S of E, ix = 2.20 ft right of O, iy = 1.38 ft above O)
  3. Determine completely the resultant of the forces acting on the step pulley shown in figure. (Ans. F = 1254.89 lb, 44.21° S of E)
  1. A 4.80-m-long beam is subjected to the forces shown. Reduce the given system of forces to (a) an equivalent force-couple system at A, (b) an equivalent force-couple system at B, (c) a single force or resultant. Note: Since the reactions at the supports are not included in the given system of forces, the given system will not maintain the beam in equilibrium. (Ans. a. 1880 Nm CW, b. 1000 Nm CCW, c. 3.13 m right of A)