6. Locate the center of mass, COM, on the bodies below then label all of the forces acting on the body in these two positions. Note: the force of gravity will be directed straight down and is located at the center of mass. Rotating clockwise (unstable) Tha Not Rotating (stable) Torque, t, is a "rotational" quantity defined by a simple equation t = Fxd, where both F and d must be perpendicular to each other. Sometimes it is advantageous to find the force component that is perpendicular to d (the distance to the pivot point). Other times it is easier to leave the force and find the perpendicular distance instead. On the diagram above extend each force of gravity vector so that it reaches the floor. The "pivot" will be the toes. The perpendicular distance will be the distance along the floor between the toes and the extended force of gravity. Label these distances on the diagram. Using these labels, explain why the person on the left rotates clockwise while the person on the right experiences no rotation (you may have to adjust your center of mass position).

6. Locate the center of mass, COM, on the bodies below then label all of the forces acting on the body in these two positions. Note: the force of gravity will be directed straight down and is located at the center of mass. Rotating clockwise (unstable) Tha Not Rotating (stable) Torque, t, is a "rotational" quantity defined by a simple equation t = Fxd, where both F and d must be perpendicular to each other. Sometimes it is advantageous to find the force component that is perpendicular to d (the distance to the pivot point). Other times it is easier to leave the force and find the perpendicular distance instead. On the diagram above extend each force of gravity vector so that it reaches the floor. The "pivot" will be the toes. The perpendicular distance will be the distance along the floor between the toes and the extended force of gravity. Label these distances on the diagram. Using these labels, explain why the person on the left rotates clockwise while the person on the right experiences no rotation (you may have to adjust your center of mass position).

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Problem 1d: Find the resultant for this system of forces: 7 ft F₁-81 lb F₁-100 lb 3 ft 40%
Choose the following systems that are equivalent to the one shown. Choose all that apply. (Remember conditions for equivalency require the sum of forces to be equal and the sum of moments about on arbitrary point to be equal) 20 lb 50 lb-ft 10 lb OA. B. OC. OD. 2 ft 90 lb-ft 20 lb 2 ft 30 lb 20 lb 2 ft 30 lb 20 lb 2 ft 2 ft 2 ft 5 lb 30 lb 2 ft 30 lb-ft 2 ft 30 lb-ft 2 ft 15 lb 2 ft
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.
Let us say that we have a ladder with its own length (L) and its own mass (m) this ladder just so happens to make a (θ) degree angle with a wall. The center of mass of this ladder is 1/3 of the way up from the bottom (it gets narrower towards the top). A person of mass M then decides to climb the ladder until he comes to the halfway point. Wall has no friction but the ground has on the ladder.. a) Draw this situation and label forces so you may set up the statics equations. b) Find the force on the ladder due to the ground (both x and y components) and the force on the ladder due to the wall.
ENTROIDS AND CENTER OF GRAVITY: The following problems are to locate centroids of areas and lines. PROBLEM 2 ONLY PROVIDE THE SAME PROCESS OF SOLUTION IN THE GIVEN EXAMPLE ? THANK YOU.
Based on the newton's second low of motion, the mass is the force applied on a system Ture O False O
If the object below has a uniform mass distribution, a=4 m, b = 8 m and c = 16 m, find %3D the centre of gravity of the object in the sketch below is %3D ) m
Find the equivalent mass (equation) if it were placed at point B.
The system of the collinear and the parallel force for the calculations of the equations in the detemination of the fluid pressure are simplified as A. The simplified collinear force system gives us a net force and the parallel force system gives us a simplified force, and then we add it vectorially B. Such simplification is not possible C. The simplification is usually done by not considering the directions of the both D. The simplification is done by considering the rotations only Select one: O A The Pascal law states that liquid at rest applies pressure at a point is in all directions. A. the same B. not the same C. not matching D. matching but not equal Select one: B D. The magnitude of the pressure of the fluid is not dependent on the. A. force B. area C. specific weight D. temperature Select one: O A B B. בם O O O
Let it be required to find the resultant of the four concurrentforces P1, P2, P3, and P4 shown in Figure 1.9a. This diagram iscalled the space diagram; it shows the relative positions of the forcesin a given system.
Consider the robotic arm shown below, made up of individual arms AB and BC. The weight of the arms AB and BC are each 160 N, and the resultant weight force is shown to act through each arm's center of gravity as shown. You may assume that the center of gravity is occurring along the midpoint of each arm's length. The robotic arm is picking up a package that weighs 40 N, and the 40 N weight force acts through the package's center of gravity at the location shown. ०८ b) 600 mm 160 N Bo 40° -600 mm 160 N 150 mm 20⁰ 40 N solve the following engineering physics problems with detailed explanations of every step. a) Consider the robotic arm AB. Calculate the moment arm of the individual arm's weight force with respect to the robotic arm's shoulder at point A. Report your answer in mm. Consider the robotic arm BC. Calculate the moment arm of the individual arm's weight force with respect to the robotic arm's shoulder at point A. Report your answer in mm. c) Consider the 40 N package. Calculate…
Consider the robotic arm shown below, made up of individual arms AB and BC. The weight of the arms AB and BC are each 160 N, and the resultant weight force is shown to act through each arm's center of gravity as shown. You may assume that the center of gravity is occurring along the midpoint of each arm's length. The robotic arm is picking up a package that weighs 40 N, and the 40 N weight force acts through the package's center of gravity at the location shown. COA 600 mm d) 160 N Bo 40° 600 mm 160 N 150 mm 20⁰ 40 N solve the following engineering physics problems with detailed explanations of every step. a) A servomotor at point B is exerting a torque value to resist the rotation induced by the weight force of the robotic arm BC and the weight force of the 40 N package. Calculate the required torque at point B necessary to resist the moment caused by the robotic arm BC and the 40 N package. Indicate the direction of the required torque ( or U). Report your answer in N-m. b) A…