x 800 lb 30 ID: 455334
Download Pdf The PPT/PDF document "The three tension forces act along the c..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
The three tension forces act along the cables. Sum the forces on a vector polygon using the x 800 lb 30° 30° AC Start/End Chapter 4: Forces in Structures and Machines © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Problem P4.18: Cable of the boom truck is hoisting the 2500lb section of precast concrete. A second cable is under tension , and workers use it to pull and adjust the position of the concrete as it is being raised, (a) Draw a free body diagram of hook treating it as a particle, (b) Determine The two tension forces act along the cables. Sum the forces on a vector polygon using the headtotail rule. Assuming the hook is in static equilibrium, the points are the same since there is zero resultant force acting on the hook. The weights of the and by applying the law of sines (Equation B.17 in Appendix B) to the vector polygon. The polygon is a graphical way of writing oo60sinlb 2500115sinAB = 2616 lb oo60sinlb 25005sinP P = 251.6 lb AB = 2616 lb and P = 252 lb It makes sense that the tension in is much larger than because weight of the concrete section. The angle between and in the vector polygon can also be found by (90 - 5 - 25. When workers are positioning the concrete, the hook may not be in static equilibrium, but may be accelerating or decelerating. In this case, the 2500 lb P 25° 2500 lb 30° 25° Start/End 5° 180° 115° 60° Chapter 4: Forces in Structures and Machines © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. The front loader with mass 4.5 Mg is shown in sideview as it lifts a 0.75 Mg load of gravel, (a) Draw a free body diagram of the front loader, (b) Determine the contact forces between the wheels and the ground, (c) How heavy a load can be carried before the loader will start to tip about its front wheels? and be the contact forces on one front and one rear tire. ThAssuming the system is in static equilibrium, write a force balance in the vertical direction and a moment balance about the front wheels. (a) See sketch above for the FBD (b) The two unknowns are 3577N 7357sm .819Mgkg 0001Mg 75021..w = 0 to eliminate + (7.357 kN) (2.0 m) (44.15 kN) (1.8 m) + (2) (3.4 m) = 0 Substitute for Rear wheel: 9.52 kN w 2R Chapter 4: Forces in Structures and Machines © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.(c) The two unknowns are and . At the condition of tipping, = 0. Apply = 0 to eliminate (2.0 m) (44.15 kN)(1.8 m)=0 The front contact force is larger than the rear because of the gravel load positioned at the front of the loader. If the load is lowered, creating a longer moment arm, then the required load to cause the loader to