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A worker applies the forces
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Chapter 3 Solutions
International Edition---engineering Mechanics: Statics, 4th Edition
- Find the internal force systems acting on sections 1 and 2.arrow_forward3. The rectangular block is to be pulled out of the ground using two strings A and B. String A is subjected to a force of 600 lb and is directed 60 degrees with the +x- axis. If the net force acting on the block is to be 1200 Ib, vertically upward, solve for the force T in string B and its direction angle 0. 600 Ib 60°arrow_forwardThe radius of the member is r= 3.45 ft and F has magnitude F= 490 lb. In two dimensions, the moment of a force can be calculated using the scalar method, MO= Fd, where F is the magnitude of the force and d is the perpendicular distance from the line of fore to the point where the moment is being considered. Using scalar method, calculate the moment about O due to Farrow_forward
- If you know that the moment of the force F about point A should look like the formula: M, i+ M,j• M, k, what is the value of the M,? knowing that the force vector equals to: F= -160.11 + -624.2) + 867Ak lb Note: Your answer should contain only numbers, not units. F B 8 ft 1.5 ft 0.25 ftarrow_forwardDifficult Problem: If magnitude of MA is 1200 kN – m and the distance between A and B is 3 m, what is the force FB and distance d so that the system shown at the right is equivalent to a force-couple system acting at point D, where the horizontal component of this force is R = 112 kN to the right and the corresponding couple is 635 counterclockwise. 60 kN 30 50 kN 2 m MA 60° 15° 200 kN m 100 kN 4 m Fs 2 m 200 kN 30 60 kN -3 marrow_forwardThe forces acting on the arm of an athlete doing shoulder exercises are shown in the figure. The athlete holds his arm at an angle of β=15° with the horizontal axis. Point O represents the axis of rotation in the shoulder joint, point A represents the connection point of the deltoid muscle to the humerus, point B represents the center of gravity of the arm, and point C represents the application point of the force acting on the hand. The distances between the rotation axis (O point) of the shoulder joint and the points A, B and C are a=20 cm, b =33 cm and c= 65 cm, respectively. The athlete's hand is exerted by a force of F= 150 N, which makes an angle of γ=30° with the vertical from the point C. The weight of the athlete's arm is W= 100 N. The direction of application of the Fm muscle force makes an angle of α=25° with the longitudinal axis of the arm. According to this; a) Calculate the FM muscle strength. b) Calculate the angle θ of the FJ joint reaction force with the horizontal.…arrow_forward
- The eyebolt shown in figure below is intended to hold several cables which exert the known forces shown. Calculate the resultant and its direction of these .forces G =4 kN 20° F = 3 kN 40° 30° 40° H = 5 kN K =2 kN R = 0.333 kN and 0 = 32 Degree R = 0.222 kN and 0 = 22 Degree R = 0.444 kN and 0 = 42 Degreearrow_forwardThe forces acting on the arm of an athlete doing shoulder exercises are shown in the figure. The athlete holds his arm at an angle of β = 30° with the horizontal axis . Point O represents the axis of rotation in the shoulder joint, point A represents the connection point of the deltoid muscle to the humerus, point B represents the center of gravity of the arm, and point C represents the application point of the force acting on the hand. The distances between the rotation axis (O point) of the shoulder joint and the points A, B and C are a=22 cm , b =33 cm and c= 60 cm , respectively . A force of F= 120 N, which makes an angle of γ=30° with the vertical from the point C, acts on the athlete's hand . Weight of the athlete's armW= 80 N. The direction of application of the Fm muscle force makes an angle of α=20° with the longitudinal axis of the arm. According to this; a) Calculate the muscle strength F M b) Calculate the angle θ of the joint reaction force F J with the horizontal.…arrow_forwardFind the resultant force (magnitude, direction and position) of a parallel force system. F1 = 40 lb up at 2 ft from point A; F2 = 70 lb down at 8 ft from point A; and F3 = 60 lb up at 13 ft from point A. Note: point A is the left end of the bararrow_forward
- The rectangular framework is subjected to the following non-concurrent system of forces: a 30lb- force below the origin, a 20 lb-force,3ft above the origin and making an angle of30°N of E, a 100lb-force with a horizontal distance of 3ft and a vertical distance of 4ftfrom the origin and making an angle of 45°NW , a 60lb -force with the same distance like the previous force but along the positive x-axis, and a 100lb- force 4ft tothe right of the origin and making an angle of 60° S of W . Determine the magnitude and direction of the resultant, as well as its moment arm relative to the origin.arrow_forwardProblem 2-16 Resolve the force Fzinto components acting along the u and v axes and detemine the magnitudes of the components. Given: F1 = 250 N F2 = 150 N e1 = 30 deg 82 = 30 deg 8z = 105 degarrow_forwardConsider the forces acting on the handle of the wrench in. Part A Determine the moment of force F₁ = {-2i +4j 8k} lb about the a axis. Express your answer in pound-inches to three significant figures. ANSWER: (Ma)1 = Part B Determine the moment of force F2 = lb. in. {3i + 7j - 6k} lb about the a axis. Express your answer in pound-inches to three significant figures. ANSWER: (Ma)2 = B lb.in. a 45° 3.5 in. in. 6 in.arrow_forward
- International Edition---engineering Mechanics: St...Mechanical EngineeringISBN:9781305501607Author:Andrew Pytel And Jaan KiusalaasPublisher:CENGAGE L