PHYS 2001 Lecture : Chap 5 Notes

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15 Mar 2019
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When applying newton"s 2nd law ((cid:4)(cid:5)(cid:6)(cid:7)(cid:8) (cid:12) (cid:13)(cid:11)(cid:5)), you should always check to see whether the problem is an equilibrium or a non-equilibrium problem. Equilibrium: (cid:4)(cid:5)(cid:6)(cid:7)(cid:8) (cid:12) (cid:10) (so (cid:11)(cid:5) (cid:12) (cid:10)) Non-equilibrium: (cid:4)(cid:5)(cid:6)(cid:7)(cid:8) (cid:9) (cid:10) (so (cid:11)(cid:5) (cid:12) (cid:10)) (cid:11)(cid:5) (cid:9) (cid:10) (cid:15)(cid:5) (cid:12) (cid:10) (cid:14)(cid:15)(cid:5) (cid:12) (cid:10) (cid:11)(cid:5) (cid:9) (cid:10) : the steel i-beam in the drawing has a weight (cid:16) (cid:12) (cid:17)(cid:18)(cid:10) kn and is being lifted upward at a constant velocity. Hint: is this an equilibrium or non-equilibrium problem? y x. Below, a concrete block is resting on a table. The table exerts a contact force on the block that is perpendicular to the contact interface between the table and the block. Here, (cid:4)(cid:6) directly opposes the block"s weight ((cid:16)). (cid:4)(cid:5)(cid:6) (cid:19) (cid:16) (cid:12) (cid:10) Note: this is not necessarily the case if there are other forces or if the block is on a non-horizontal surfaces. This is a free body diagram: (cid:13) (cid:24)

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