The force acting on the gear tooth is F=45 KN. Resolve this force into two components acting along the lines aa and bb a 80° 50° b.
Q: Q1. The 300lb force acts on the box B. Resolve this force into two components, one along AO and the…
A: Free body Diagram
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Q: Q1: Resolve the force F2 into components acting along the u and v axes.
A: Given data F1 = 300 N F2 = 500 N
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Q: 4. The pole is subjected to the force F which has components F= 2.5 kN and F= 1.5 kN. If B = 60°,…
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Q: If F_1=125N, F_2=25N and F_3=150N, determine the resultant force in t
A: Given Data : F1=125 NF2=25 NF3=150 N
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Q: The force acting on the gear tooth is F=55 KN. Resolve this force into two components acting along…
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Q: Q1: If F, - 150 lb and e- 55', determine the magnitude and direction measured clockwise from the…
A: As per policy, first question is answered in detail. Second question can be posted as separate…
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Q: F2–10. If the resultant force acting on the bracket is to be 750 N directed along the positive x…
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Q: H.W1:The 624 N force is a resultant of two force P & Q shown in fig. determine these forces . 624 N…
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A: Given Force F= (240N)j - (680 N)k To find Magnitude and direction
Q: Determine the resultant force acting at A. 1.5 m 2 m 4 m F= 40 N F=55 N 6 m 3 m 4m.
A: Given: The force along the AB is, FB=55 N. The force along the AC is, FC=40 N.…
Q: The force acting on the gear tooth is F=80 KN. Resolve this force into two components acting along…
A: The force acting on the gear tooth is, F=80 kN.
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Q: Q1 ) The R is resultant of three forces (F1, F2 & F3 ), determine the magnitude and direction of F3…
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Q: 30° Fz = 200 N 1213 F3= 260 N If the magnitude of the resultant force acting on the bracket is to be…
A: Write the given vectors in the form of vectors. F2→=200i∧F3→=260513i∧-1213j∧…
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Q: The force acting on the gear tooth is F = 20 lb. Resolve this force into two components acting along…
A: Free body diagram FREE BODY DIAGRAM is a diagram of the system which represents all the forces…
Q: H.W1:The 624 N force is a resultant of two force P & Q shown in fig. determine these forces. 624 N 3…
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Q: Q1: The 650 N tension force in cable AB is required to prevent the CA bar from rotating clockwise.…
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Q: If F 500 N and Fe 800 N, determine the magnitude and coordinate direction angles of the resultant…
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Q: Q1. The 300lb force acts on the box B. Resolve this force into two components, one along AO and the…
A: Consider the box shown and as it is asking for the resolving the 300 lb force, it can be inferred…
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Q: Q1(30%) The R is resultant of three forces (F1, F2 & F3 ), determine the magnitude and direction of…
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Q: Q2: The two cables have the same tension T and sum to produce a force T. of 1000 Ib on the pulley.…
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Q: Q2: Determine the magnitude of the force (F) so that the resultant ( FR ) of the three forces is as…
A: The provided force system is shown below.
Q: The forces F1, F2, and F3, all of which act on point A of the bracket. Determine the F1 and the…
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Q: Q2) Resolve the force F2 into components acting along the u and v axes and determine the magnitudes…
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Q: Q11/ Determine the magnitude of the resultant force acting on the plate and its direction, measured…
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Q: Determine the magnitude of the resultant force (R) acting on the plate and its direction (e)…
A: Answer: (a)The magnitude of the resultant force is 730.92 N. (b) It's direction measured from…
Q: F
A: Given Data⇒ Resultant forceFR=-163i+619j+136k
Q: Q1: The 700 N tension force in cable AB is required to prevent the CA bar from rotating clockwise.…
A: Solution: The force can be resolved as Fx and Fy as shown in the figure (Fx and Fy are…
Q: Q3. The 500 N force acting on the frame is to be resolved into two components acting along the axis…
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Q: Q1: The 800 N tension force in cable AB is required to prevent the CA bar from rotating clockwise.…
A: Draw the free body diagram. AD=1.5 msin 70°AD=1.41 m CD=1.5 mcos 70°CD=0.51 m…
Q: Determine the magnitude and direction of the force F= (320 N)j + (400 N)j - (250 N)k.
A: F= (320 N)j + (400 N)j - (250 N)k
Q: If F₁ = 215 N and d = 46°, determine the magnitude of the resultant force acting on the bracket. 30°…
A: Given Force, F1 = 215 N F2 = 200 N F3 = 260 N Find Resultant force
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- resultant force on the contracting gear tooth issolve show all steps/solutions show fbd with all forces and moments/torquesFour forces, FA, FB, FC, and FD, each of magnitude 1 N, are applied to a square object at the points shown below (at the tails of the arrows), and each produces a torque around the pivot (the circle). Rank these torques, τA, τB, τC , and τD, from most negative to most positive. Use the sign convention for torques. a)τA < τD < τC < τB b)τA < τD < τB < τC c)τC < τD < τB < τA d)τB < τC < τA < τD e)τD < τB < τA < τC
- 3. At temperature T₁, the two rigidly connected rods just fit the rigid supports. If both sections are identical except for the diameter, prove that the following equations are true when temperature rises to T₂. a. F= b. OAB C. OBC A x(T₂-T₁)лd²E 20 x(T₂-T₁)E 5 9x(T₂-T₁)E 5 -1/2- d B 1/12- d CIn (Figure 1), take 0 = 30°. Figure B 0 -82 1 of 1 400 lb Part A Determine the design angle 0° ≤ ≤ 90° between members AB and AC so that the 400-lb horizontal force has a component of 650 lb which acts up to the right, in the direction from B toward A. Express your answer in degrees to three significant figures. ø= Submit Part B Submit AΣ vec Request Answer Determine the magnitude of the force component along AC. Express your answer in pounds to three significant figures. FAC = 408 Provide Feedback VAΣ vec Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining www ? ? lb Next >In Exercises 25 and 26, find the magnitude of the torque exerted by F on the bolt at P if PQ = pounds. 8 in. and F = 30 lb. Answer in foot- %3D 25. 26. 135° F 60°
- F2 F1 Figure (4) 200 For figure (4): The ring is subjected to two forces (F1 and F2). Suppose that the resultant force has the value of 225 N and it is located downward in the third quart at angle 40° measured from the y-axis. Set 0 = 30° , ß = 80°. em 11- What is the value of the force F1? 1312 - A00 A= 50° B=70%This shoulder bolt is composed of hot-rolled monel. The head of the shoulder bolt is a regular hexagon and it measures 6 mm across opposite flats. The other dimensions of the bolt are given below, as well as the forces to which this component is subjected. Find the change in length of each segment of the part as well as the overall change in length from A to D F_A = 4.8 kN F_B = 5.1 kN F_C =1.6 kN F_D = 1.9 kN a = 4mm b = 4.1 cm c = 6.8 cm d_s = 5 mm d_t (effective) = 3mm Assume:∙ Bearing stress between threads of the bolt and the nut may be neglected∙ Stress concentration effects may be neglected∙ The force that the nut applies may be treated as a concentrated force.∙ The nut may be treated as rigid.∙ The forces applied to the nut and the shoulders of the bolt are distributed evenly around their perimeters δAB = δBC = δCD = δoverall =0/10 Question 6 of 10 Four forces are exerted on the eyebolt as shown. If the net effect on the bolt is a direct pull of 550 lb in the y-direction, determine the necessary values of T and 8. Assume F1380 lb, F2=245 lb, F3=420 lb, 中 = 28 $ F3 F. Answers: T= i e= 요 r ! lb ! ||| Co
- 1-63.M. The key in Figure 1-18 has the dimensions b = 10 min, h = 8 nim, and L = 22 min. Determine the shear stress in the key when 95 N-m of torque is transferred from the 35-mm-diameter shaft to the hub. Hub End view Shear plane Shear pl T=Torque -D=Shaft diameter T=F(D/2) F₁-Force of shaf F₂= Force of hub on 1 Shear area=A,=bxLP1=2bar X1=o.41. Four pulleys are attached to the 55 mm diameter steel shaft. If torques is applied to the pulleys as shown in the figure, determine the angle of rotation of pulley D relative to pulley A. 2300 N-m 1200 N-m 2000 N-m 1500 N-m C D 4 m 2 m