{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

notes11a - Static Equilibrium In Chap 6 we studied the...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
In Chap. 6 we studied the equilibrium of point- objects (mass m) with the application of Newton’s Laws Therefore, no linear (translational) acceleration, a =0 = = 0 , 0 y x F F Static Equilibrium
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
For rigid bodies (non-point-like 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, α =0 (don’t need to know moment of inertia) We can now use these three relations to solve problems for rigid bodies in equilibrium ( a =0, α =0) Example Problem The wheels, axle, and handles of a wheelbarrow weigh 60.0 N. The load chamber and its contents = 0 τ
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 ( 0 0 0 3 2 1 3 2 1 = + = + = = + = + + = L L W D L L W D L W D F F L L F L F F L F L F
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}