Newton’s second law7

Newton’s second law7 - 12.005 Lecture Notes...

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12.005 Lecture Notes 7 Newton’s second law For a point mass Fm a = ± ± ± ± ± We can obtain from a free-body diagram – e.g., pendulum. F Figure 7.1 Both and are vectors. F a F i = ma i i = 1, 2, 3 For a continuum, we can also construct a free-body diagram. It’s easiest to do this component-by- component. F ~ T ~ mg ~ Figure by MIT OCW.
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Consider the following figure. Figure 7.2 Consider forces on faces. face Traction ( T 3 ) area left σ 13 2 3 dx dx right 13 + 13 x 1 dx 1 dx dx 2 3 back 23 1 3 dx dx front 23 + 23 x 2 dx 2 dx dx 1 3 top 33 1 2 dx dx bottom 33 + 33 x 3 dx 3 dx dx 1 2 Force ( F 3 ) in x 1 direction: 13 x 1 dx 1 dx 2 dx 3 σ 33 σ 13 + σ 13 x 2 x 3 x 1 dx 1 x 1 σ 33 + σ 33 3 x 3 Figure by MIT OCW.
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Force ( F 3 ) in x 2 direction: σ 23 x 2 dx 1 dx 2 dx 3 Force ( F 3 ) in x 3 direction: 33 x 3 dx 1 dx 2 dx 3 Body force: ρ b 3 dx 1 dx 2 dx 3 Combining 13 x 1 dx 1 dx 2 dx 3 + 23 x 2 dx 1 dx 2 dx 3 + 33 x 3 dx 1 dx 2 dx 3 + b 3 dx 1 dx 2 dx 3 = a 3 dx 1 dx 2 dx 3 Dividing through by δ V = dx 1 dx 2 dx 3 13 x 1 + 23 x 2 + 33 x 3 + b 3 = a 3 Similar analysis gives
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This note was uploaded on 10/25/2010 for the course MIT Geodynamic taught by Professor Ywn during the Fall '10 term at MIT.

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Newton’s second law7 - 12.005 Lecture Notes...

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