Answered: A force of 500 N is applied on a rigid…

A force of 500 N is applied on a rigid bar as shown in the figure The moment of Fabout origin O in clockwise direction is F 4 m 60° Select one O a 2732 N m O 1732 Nm Oc 1000 Nm Oa. 732 N.m

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter9: Moments And Products Of Inertia Of Areas

Section: Chapter Questions

Problem 9.13P: Figure (a) shows the cross section of a column that uses a structural shape known as W867...

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Figure (a) shows the cross section of a column that uses a structural shape known as W867 (wide-flange beam, nominally 8 in. deep, weighing 67 lb/ft). The American Institute of Steel Construction Structural Steel Handbook lists the following cross-sectional properties: A=19.7in.2,Ix=272in.4, and Iy=88.6in.4. Determine the dimensions of the rectangle in Fig. (b) that has the same Ix and Iy as a W867 section.
The bent rod of negligible weight is supported by the ball-and-socket joint at B and the cables attached to points A and C. Find the forces in the cables and the magnitude of the reaction at B. Dimensions Figure P.5.39
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Topic: Couples, Resultant of Force Systems & Equilibrium If F1 = 10N, F2 = 15N, F3 = 8N, F4 = 10N, F5 = 18N, and the length of te side of a red square is 1m. Calculate magnltude (+ for CCW, - for CW) of the resuItant moment (N-m) about point A. Calculate the magnitude of the reItant f0rce (N). (Answer: 50.80Nm, 20.33N)
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Consider the image below: A horizontal beam of length 8 m is pinned at its left end so that it is free to rotate about that end. It has a nonuniform linear mass density given by λ(x) = 78 x (kg/m). Its right end is secured by a rope, the angle shown being 45 degrees. Calculate the magnitude of the tension in the rope, in N. (Please answer to the fourth decimal place - i.e 14.3225)
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 downw
The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and cast is W=180 N. The horizontal distance between points A and B where the cables are attached to the leg is L=100 cm and the vertical distance is d=5 cm. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=30°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N) Answerb) Find the tensile force T2 in cable 2. (Write your result in N) Answerc) Find the angle α of cable 1 with the horizontal. Response
The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and the cast is W=210 N. The horizontal distance between points A and B where the cables are attached to the leg is L=100 cm and the vertical distance is d=6 cm. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=33°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N) b) Find the tensile force T2 in cable 2. (Write your result in N) c) Find the angle α of cable 1 with the horizontal.
The bar AC is supported by a joint at A and a pin B that is free to slide in the same direction (alpha) of the bar. A bar is connected at C and a force is applied in the theta direction. Consider that L1 = 2.7 m, L2 = 1.7 m, alpha = 13 °, theta = 25 °, and P = 68 kN. Determine the magnitude of the force (kN) on pin B.
The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and cast is W=180 N. The distance between the points A and B where the cables are attached to the leg is given as L=100 cm and the angle of the leg with the horizontal is given as γ=8°. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=50°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N)b) Find the tensile force T2 in cable 2. (Write your result in N)c) Find the angle α of cable 1 with the horizontal.
A sign of weight 2280 N is supported as shown by a ball and socket support at A and two cables BC and AD 1- Tension in cable BC is (N) : (a. 4560 - b. 6840 - c. 5700 - d. 7980 - e. 9120 ) 2- Tension in cable AD is (N): (a. 10260 - b. 13680 - c. 15960 - d. 18240 - e. 10640 ) 3-Magnitude of x-component of reaction at O is (N): (a. 5700 - b. 6840 - c. 7980 - d. 9120 - e. 10260 ) 4- Find Magnitude of y-component of reaction at O is (N) 5- Find Magnitude of z-component of reaction at O is (N)