T1a9s - Assignment 9 solutions (Fall term Physics 102) 8.62...

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Assignment 9 solutions (Fall term Physics 102) 8.62 Use an axis perpendicular to the page and passing through the lower left corner of the frame. Then, 0 τ Σ= gives () ( ) ( )( ) 1 1 10.0 N 0.150 m cos 50.0 0.150 m sin 50.0 0.300 m 0, T T −− ° +°= or 1 T = 11.2 N 0.150 m T 2 T 1 cos50.0° T 1 sin50.0° 10.0 N F 0.150 m 0.300 m Then, using , obtain 0 y F ( ) 2 11.2 N sin 50.0 10.0 N=0 T + °− , or 2 T = 1.39 N Finally, Σ gives , or = 0 x F 1 cos 50.0 0 FT −° = ( ) 11.2 N cos 50.0 F = °= 7.23 N . 9.6 From 0 FL Y AL = , the tension needed to stretch the wire by 0.10 mm is ( )( ) 2 00 4 Yd L YA L F LL π == ( ) ( ) ( ) 2 10 3 3 -2 18 10 Pa 0.22 10 m 0.10 10 m 22 N 4 3.1 10 m ××× × The tension in the wire exerts a force of magnitude F on the tooth in each direction along the length of the wire as shown in the above sketch. The resultant force exerted on the tooth has an x -component of cos 30 cos 30 0 xx RFF F = ° + ° = , and a y -component of . Thus, the resultant force is R = 2 sin 30 sin 30 22 N yy F F =− °=− =− 2 N directed down the page in the diagram .
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T1a9s - Assignment 9 solutions (Fall term Physics 102) 8.62...

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