2. Forces T1, T2, and T3 are acting on the eye bolt. If the magnitudes of T1, T2, and T3 are 330 N, 220 N, and 440 N, respectively, what is the direction of their resultant?

The resultant of the concurrent forces has a magnitude of 800 N passing through points (x₁ = 1, y₁ = 1, z₁ = 1) and (x₂ = 2, y₂ = 3, z₂ = 4). 1. Compute the x- component of the resultant force.  a. 371.1 N b. 392.31 N c. 422.6 N d. 213.81 N 2. Compute the y – component of the resultant force.  a. 512.92 N b. 556.62 N c. 427.62 N d. 634.72 N 3. Compute the z – component of the resultant force.  a. 632.73 N b. 641.43 N c. 742.13 N d. 791.43 N

Three forces act on a hook. Determine the magnitude of the resultant of the force. Neglect hook bending. F2 - 1500 N - 1000 N 20 30 16 F3 - 750 N Select one: O a. 1250 N Ob. 1510 N C. 1140 N d. 989 N

! Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Two forces P and Q are applied as shown at point A of a hook support. ट A 20°i 35⁰ Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using the parallelogram law. The magnitude of the resultant is N. The direction of the resultant considering OO' as the reference axis is

Given a set of concurrent forces with force A = 25 lbs, direction = 30 degrees; force B = 40 lbs, direction = 120 degrees. Find the magnitude and direction of the resultant force. a. 52.36 lbsb. 47.17 lbsc. 34.68 lbsd. 67.45 lbs

Express the 40 lb force in the rope from point A to point C in cartesian format. Point C is in the x-y plane and at an angle of 20 degrees from the y axis as shown. 4 ft 20 lb A B 3 ft X 40 lb Z 2 ft 1.5 ft 4.87i+27.6j - 28.5k lb -4.87i+27.6j + 28.5k lb -4.87i+27.6j - 28.5k lb 4.87i+22.8j - 32.5k lb 2 ft 20 y C

The Force F produces a moment of 3456 N⋅m clockwise at point A, “M” N⋅m counter – clockwise at point, and “M” N⋅m clockwise at the origin O. The Force F acts at point C. Determine the following: a.) The Force F. b.) The coordinate y.

PLS be legible with your work! I thought it was 2.13 and 3.81, respectively, but I think I'm missing a step. Thx!

What would be the resultant force F of the triangular distributed load and its location d measured from point A.  Select one: a. F = 300 lb and d = 1.67 ft b. F = 150 lb and d = 3.67 ft c. F = 300 lb and d = 2.5 ft d. F = 150 lb and d = 3.33 ft

2. Determine the following: a) Magnitude of the projected component of along the line of action of F2 b) Angle 0 between the two cables attached to the pipe. F, = MN Ib 120 60° F, = SN Ib 1.5 m 3. The guy wires are used to support the telephone pole. Represent the force in each wire in Cartesian vector form. Neglect the diameter of the pole F= MNN F: = FN N 4 m 2 m D. 4 m 3 m y