This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: In Chap. 6 we studied the equilibrium of point objects (mass m) with the application of Newtons Laws Therefore, no linear (translational) acceleration, a =0 = = , y x F F Static Equilibrium For rigid bodies (nonpointlike objects), we can apply another condition which describes the lack of rotational motion If the net of all the applied torques is zero, we have no rotational (angular) acceleration, =0 (dont need to know moment of inertia) We can now use these three relations to solve problems for rigid bodies in equilibrium ( a =0, =0) Example Problem The wheels, axle, and handles of a wheelbarrow weigh 60.0 N. The load chamber and its contents = weigh 525 N. It is well known that the wheel barrow is much easier to use if the center of gravity of the load is placed directly over the axle. Verify this fact by calculating the vertical lifting load required to support the wheelbarrow for the two situations shown. F L L 1 L 2 L 3 F w F D F L L 2 L 3 F w F D L 1 = 0.400 m, L 2 = 0.700 m, L 3 = 1.300 m First, draw a FBD labeling forces and lengths from the axis of rotation L 1 L 2 F w L 3 F D F L axis N 194 m 300 . 1 m) N)(0.700 60.0 ( m) N)(0.400 525 ( 3...
View
Full
Document
This note was uploaded on 12/10/2011 for the course PHY 1111 taught by Professor Stencil during the Fall '11 term at University of Georgia Athens.
 Fall '11
 Stencil
 Acceleration, Mass, Static Equilibrium

Click to edit the document details