A bent pipe is situated in the y-z plane. The force, F₁ = 26 lb (negative x-direction) is attached at the end of a bent pipe, and a moment, M = 228 ft-lb CW is applied around the center shaft (z-axis). The dimensions are a = 5.9 ft., b = 4 ft., What is the moment created around the origin.? Report your answer in cartesian form. Z a M X y < >

The moment of the force F=960-lb about the point O in Cartesian vector form is. 1 ft 4 ft 2ft Select one: O a. Mo = 1600 – 3200A O b. Mo = -200j + 400k Oc. Mo =-200; – 400k O d. Mo= 200; + 400k 1:15 PM

The force vector F = (ai + bj + ck) kN acts on point A of the system shown in the figure. Bar OB and point B with coordinates B (x, y, 0) m is on the x-y plane. By using the values given to your name in the data table;a) Find the moment at point O in vectorial form,b) Calculate the magnitude of the moment with respect to the rod OB F = {ai + bj + ck} kN A = 60 B = 70 C = 80 Meter L1 = 16 L2 = 6 B (x, y, 0) coordinates (m) x = 16 y = 3

Example: The resultant of the concurrent forces shown in figure is 300 N pointing up along the y-axis. Compute the value of F and e required to give this resultant. 50ON 30 240N o Since the resultant is up along y- axis then EF = 0 o HW. o F = Fcos 0 + 240 cos 30 – 500 = 0 o Fcos 0- 292.15 = 0 o Repeat this problem if the resultant is 40ON down to the right at 60° with the x-axis. o Also repeat the problem if the body is in equilibrium. o Fcos 0 = 292.15... .... (1) o F, = R = Fsin 0 - 240 sin 30 = 300 Fsin 0 = 300+ 120 = 420 ..... (2) Divide eqn. (2) by (1) yields F sin 0 420 %3D F cos e 292.15 o tan 0 = 1.4376, 0 = tan 1.4376 = 55.18° 420 F = sin 55.18 = 511.6N

Review Learning Goal: To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member. Part A - Moment due to a force specified by magnitude and endpoints The moment of a force F about the moment axis passing through O and perpendicular to the plane containing O and F can be expressed using the vector cross product, Mo =r x F. In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant: As shown, a member is fixed at the origin, point O, and has an applied force F, the tension in the rope, applied at the free end, point B. (Figure 1) The force has magnitude F = 180 N and is directed as shown. The dimensions are ¤1 = 0.350 m, x2 = 1.90 m, y1 = 2.30 m, and z1 = 1.20 m. What is the moment about the origin due to the applied force F? i j M =rx F =|rz k Express the individual components of the Cartesian vector to three significant figures, separated by commas. ry F F,…

This is a street sign ALLE Ľ d Cricen: 0₁ = 31.0° and Force ₁ = 16.8 N 02 = 14.5⁰ and force = 9.25 N d = 22 35.25cm Find the moment tffe street sign in (N-m) about its center experiences in (N-m)

Find unit vectors that satisfy the stated conditions. NOTE: Enter the exact answers in terms of i, j and k. (a) Same direction as - 2i + 5j. u = (b) Oppositely directed to 9i – 6j + 6k. (c) Same direction as the vector from the point A(-2,0, 4) to the point В(3, 3, 3). w =

What is the magnitude of the moment done by the force about the "z-axis" in the following figure. A penubore 4 m F= (60i – 50j – 20k} N 3m 7 m 4m 6 m

Questions Ib and le refer to the figure shown. F=|-201 + 10j + 15k) N lb. What is the moment of F about point A? (a) - 60 i +0j+ 120 k (N-m) 4 m (b) 60 i + 10 j - 120 k (N-m) (c) - 20 i + 125j-110 k (N-m) 4m (d) 20 i - 125 j+ 10 k(N-m) lc. What is the moment of F about line AB? (a) 53.64 N-m (b) 88 N-m (c) - 88 N-m (d) - 53.64 N-m

Z B Part B - Moment due to a force specified as a Cartesian vector As shown, a member is fixed at the origin, point O, and has an applied force F, the tension in the rope, applied at the free end, point B. (Figure 2) The force is given by F = 120 Ni - 120 Nj+50 N k. The dimensions are x₁ = 4.25 m, y₁ = 6.25 m, and z₁ = 2.25 m. What is the moment about the origin due to the applied force F? Express the individual components of the Cartesian vector to three significant figures, separated by commas.