Consider the bracket shown in Fig-1. If: • F1 = 10 N@ 120° • F2 = 75 N@ -90° then, what is the magnitude of the resultant force on the bracket? 8.66 N 66.5 N -75 N 75 N O ON 66.3 N*m O 100 N*m O 10 N
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- Answer for the 1st part of the problem. magnitude of force A = 74 N direction of force A = 25o north of west second force = P direction of force P = α south of west arrow_forward The resultant of two forces is horizontal , it means that the y component of resultant = 0 Ry = ASin(25o) - PSin(α) 0 = (74×0.42) - PSin(α) PSin(α) = 31.08 NFor, the smallest force P , the value of Sinα must be maximum which is 1 when the value of αis 90 degree Thus, P×Sin(90o) = 31.08 N P = 31.08 N The magnitude of smallest force P = 31.08 N the direction of force P = 90 degree south of west = along - y axis= arrow pointing vertically downwConsider the bracket shown in Fig-1. If: F1 = 10 N@ 120° • F2 = 75 N@ -90° then, what is the direction of the resultant force on the bracket? 176° 4.31° -4.31° 2.17° -176° -2.17° 94.3° -94.3°* O l 42%016:05 R The metal plate is held by two cables as shown in the following figure. If the force of each cable acting at A is FAB = 100 Ib and Fac = 200 Ib. Determine the magnitude of the resultant force of the two forces acting at A. -8 ft- 8 ft 6 ft B Y FAB FAC A(10;6;0) Select one: a. FR = 258.96 lb b. FR = 100 Ib c. FR = 200 Ib d. FR = 300 Ib Ouootion 7 II
- * O 42%À 16:04 R The support, shown in the following figure, is subjected to the action of two forces F1 = 180 N and F2 = 400 N. If a2 = 45° and B2 = 60°, determine the magnitude of the resultant force of the two forces. Z. F2 Y2 B2 60° 15° F = 180 N PROBO2_076-077.jpg Copyright 2010 Pearson Prentice Hall, Inc. Select one: a. FR = 541.95 [N] b. FR = 580 [N] c. FR = 400 [N] d. FR = 380.8 [N] IIExample ( 2-1) : The ring in the figure below is subjected to two forces ( F1) and ( F2). Determine the magnitude and direction of the resultant force. 10° F, = 150 N F = 100 N %3D 15° Solution : Parallelogram Law. The parallelogram is formed by drawing a line from the head of F1 that is parallel to F2 , and another line from the head of F2 that is parallel to F1 , the resultant force FR extends to where these lines intersect at point A. The two unknowns are the magnitude of ( FR) and the angle ( 0) (theta).Use the method of SECTIONS to find the true magnitude and direction in bar HG of the truss shown below in Fig 3.5 -500lb -100lb A 45° -200lb H -100lb 2 B -200lb -200lb MAGNITUDE AND DIRECTION OF SUPPORTS -500lb H -100lb 45° 2¹ C2D 2¹ Fig 3.5. -200lb -100lb LI 2 B 2₁ 2₁ D 2₁ E
- An eye bolt is used to attach 3 cables to a steel plate. The tension in the three cables create F,=200 Ibf, F2=250 lbf, and F3=100 lbf with 0 = 30 degrees and p=27.1 degrees. If the eye bolt is in equilibrium, what is the x-component of the sum of other forces (reaction force) on the bolt? If you add up the three force vectors, the reaction force you are looking for will just be in the opposite direction to put the eye bolt in equilibrium. The x-direction is positive and to the right. Eye bolt rsteel plate Nut & Washer2.3 EXAMPLE 2.1 The screw eye in Fig. 2-11la is subjected to two forces, F, and F. Determine the magnitude and direction of the resultant force. 10 e150 N F 100 N 15 (a) SOLUTION Parallelogram Law. The parallelogram is formed by irawing a bneP 4:01 A moodle.nct.edu.om 480% * 4G OMANTEL l. Find the magnitude and direction of the resultant force for the system of forces shown in the figure. Where F1 =15 N, F2 =35N, F3 =5 N, F4 =15N. The angles 01 =15°,02 =25° , 03 = 30°, 04 = 20 ° (Enter only the values in the boxes by referring the units given in bracket. Also upload the hand written copy in the link provided) F2 F1 02 F3 F4 II
- A transmission tower (a truss) for supporting electric wires is shown. The load of wire at D is shown as force F1 at an angle of a and the load of wire at A is shown as force F2 at an angle of B. Using the values given in the table below: F1 (N) F: (N) a (deg) B (deg) w (m) h (m) Value 150 70 30 60 2 a) Find the reaction forces at supports Y and Z. b) Determine the forces supported by members DO, DM, SV, and UX and specify whether the member is in Tension or Compression. M Q B F1 F2 T W X hTwo forces are applied to a hook. Variable F₁₂ Ą a b FR= d FR=( Q= B. values for the figure are given in the following table. Note the figure may not be to scale. value 75 lb 35 lb 2.5 ft B= Y = 4.5 ft 9 ft 1 ft Z a. write each force (F₁,F) as cartesian vectors. b. Find the resultant force FR express as a cartesian vector c. Find the magnitude of the resultant force FR i. a ii. B iii. Y Round your final answers to 3 significant digits/figures. Submit Question → d. Determine the coordinate direction angles of the resultant force FR lb 13/12 a degrees F₂ degrees F₁ degrees A A) lb y e) lb AlbConsider the bracket shown in Fig-3. If: • F1 = 10 lb @ -90° • a - 2 ft • b = 4 ft • c= 1 ft then, what is the magnitude of the internal force within member 'BD'? O lb 10 lb 14.2 Ib 20 lb 26.4 lb 30 lb 32.3 lb 40 Ib 45 lb 50 Ib 54.1 Ib 60 Ib 67.1 lb 70 lb 72.7 lb