250 N 400 N 30° 300 N Q 1: For the figure above, calculate: A. Sum of resultant force on (x – axis) N) B. direction (→ +) C. Sum of resultant force on (y – axis) N)
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- Determine the magnitude and direction of the resultant force using parallelogram law of forces for the figure given below.Consider the following coplanar system of tensile forces: F1 = 90 N at 180°, F2 = 150 N at 310° and Fy = 400 N at 0°. Use the counter clockwise direction as the positive direction of the angle from the positive x-axis. Provide the answers to 2 decimal places.Task 2 Determine, using complex numbers, the magnitude and direction of the resultant of the coplanar forces given below, which are acting at a point. Force A, 5 N acting horizontally, Force B, 9 N acting at an angle of 135◦ to force A, Force C, 12 N acting at an angle of 240◦ to force A.
- Expiremental Method: Force Table F1 - 150N, 30° N of E F2 - 100N, 60° E of S 1) Find the Equilibrant Magnitude and Equilibrant Direction 2) Find Resultant Force Vector Magnitude and %Error 3) Find Resultant Force Vector Direction and %ErrorFind the x-component a force, 100N, acting 60 degrees below the x-axis,Determine the internal normal force at point C. Take F1F1F_1 = 460 lblb , F2F2F_2 = 250 lblb , F3F3F_3 = 346 lblb . Determine the shear force at point C. Determine the moment at point C. Determine the internal normal force at point D. Determine the moment at point D.
- We have 2 individuals pulling a marshmellow with ropes. The resultant force is directed along the positive x-axis. Person A is pulling at 60 degree angle. 1. Find what angle ? is if individual B is pulling twice as hard as individual A. 2. Draw the vector addition relationship between the force vectors ??, ??. Also, find FA and FB resultant force, shown using ?. Use either the parallelogram law or the triangle rule.The following concurrent forces are acting on a body in xy plane: A 300 lb pointing up to the right 30 degrees with x axis, 400 lb at 45 degrees with x axis acting up to the left, 200 lb down to the left at 60 degrees with x axis, 150 lb down to the right at 75 degrees with x axis and a 100 lb directly to the right. Determine the resultant force and its direction using graphical and analytical method. Pls show the solution, formula and free body diagram.Given uAF=1.5/6.5i+2/6.5j−6/6.5k, uAE=−1/3i−2/3j−2/3k, uAG=2/7i+3/7j−6/7k, and magnitude of the force FAF=1083 N. Find the projection on line AE of the force in cable AF.
- Find the resultant force for the forces shown below.Assuming F1 = 150 N and F2 = 200 N.In the figure below, all the bearings are movable bearings and the C bearing is in an inclined position. Draw the M, N, T diagrams according to your force values. (20kN force is perpendicular to the beam it acts on. P1, P2, P3 forces are in the vertical direction.) q=7 kN/m P1=25 kN P2=27 kN P3=11 kNIf the component of a force along x-axis is Fx = 20 kN and along y-axis is Fy = 30 kN, and if “t” is an axis making an angle of 30o in the counter clockwise direction with respect to x-axis, what would be the components of this force Ft and Fn along the orthogonal axes “t” and “n”, respectively? Select one: Ft = 2.32 kN and Fn = 35.98 kN Ft = 15.98 kN and Fn = 32.32 kN Ft = 35.98 kN and Fn = 2.32 kN Ft = 32.32 kN and Fn =15.98 kN