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## Phys for Science & Engr I

by: Cale Ondricka

22

0

4

# Phys for Science & Engr I PHYS 2101

Cale Ondricka
UNCC
GPA 3.94

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COURSE
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## Popular in Physics 2

This 4 page Class Notes was uploaded by Cale Ondricka on Sunday October 25, 2015. The Class Notes belongs to PHYS 2101 at University of North Carolina - Charlotte taught by Awad Gerges in Fall. Since its upload, it has received 22 views. For similar materials see /class/229009/phys-2101-university-of-north-carolina-charlotte in Physics 2 at University of North Carolina - Charlotte.

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Date Created: 10/25/15
Physics 2101 Static Equilibrium Instructions 1 Read your text book Chapter 1 1 Sections 11 112 and 113 Eg uilibrium of a particle and a rigid bodv According to Newton s second law a particle is at equilibrium if lt gt This gives Fnetxzo Fnetho This condition is enough because the particle has a very small radius that tends to zero as shown in the gure However this condition is not enough for a rigid body ofextended dimensions For example consider a rod with two equal and opposite forces applied to it as shown in the next gure Even though the net force on the rod is zero it will rotate around an axis perpendicular to the plane ofthe paper that passes through the pivot point 0 lt r24v O F A rigid body is said to be in equilibrium when it is in translational and rotational equilibrium that is m 3 Translational equilibrium AND 3 Rotational equilibrium Example The rod shown in the figure is at equilibrium Find the magnitudes ofthe tension forces T1 T2 and T3 Tlsine Tlcose Newton s equations give F 7cos40 7 0 1 mm Fm TI sin 40 T2 700N mg 0 2 We need another equation because we have 3 unknown quantities We will take the torque with respect with an axis perpendicular to the plane of the page and passing through 0 In this way the torque due to T2 and T3 will cancel out since they pass by the pivot 0 d 0 Thus rm 0 700N05m mg lmTI 2msin 40 700N05m mglm 3 TI E 5 ON 2m sm 40 From equation 1 3 T 00540 T3 0 3 T3 T 00540 383N From equation 2 3 T2 700N 30kg 98 TI sin 40 673N s Stabili of Structures Potential energy stored on an object is given by the limited integral of the conservative force over the psth moved by the object AU UX UM W j Fxdx x03 X0 As an opposite operation if the potential distribution is known as a function of position Uxyz the components of forces can be calculated using the partial derivative of potential function as FX 6Uxyz F 6Uxyz F 6Uxyz 6x 9 y 9 2 62 Equilibrium occurs when FX 2 F F 0 For one dimensional problems The potential energy distribution is a function of x only and the force FX is defined as FX 2 d Ux dx And equilibrium occurs when FX 2 d U x 2 d2 0 There are three different cases of stability 1 Stable Eguilibrium 2 d 0a gt0 Atx x0 dx2 This means that U has a minimum at x X0gt If the object is displaced from x0 a restoring force is created to return it back to KC Ux x 2 Unstable Eguilibrium 2 d UQ lt0 Atxx0 dx2 This means that U has a maximum at x x0gt If the object is displaced from x0 a force is created to drive it away from x0 3 Neutral Eguilibrium Atxx0 dx2 d2UX 0 This means that U is a constant function If the object is displaced from x0 it will not return back nor move away but will still be at equilibrium 7 um ELQQD LUCK

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