ME211solution_1 - (a) Fy sin 45 = - Fn 20 = sin 75 sin 60...

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(a) 20 sin 45 sin 75 sin 60 y n F F == °°° 16.3 lb 22.3 lb y n F F = =− (b) 20 sin15 sin 45 sin120 tx FF °° ° 5.98 lb 16.3 lb t x F F = =
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Express each force in Cartesian vector form ( sin 30 ) ( cos30 ) AA FF + A Fi ° j ( cos 20 ) ( sin 20 ) BB =− ° + ° B j Sum two forces to get resultant force ( sin 30 cos 20 ) ( cos30 sin 20 ) AB A B Rx Ry F F =+ ° + ° + RAB FFF i ° j ij 700sin 30 600cos 20 213.8 N Rx F − ° = 700cos30 600sin 20 811.4 N Ry F = The magnitude of the resultant force is 22 2 2 ( 213.8) 811.4 839 N RR x R y F = + = The directional angle measured counterclockwise from positive y axis is 11 213.8 tan tan 14.8 811.4 Rx Ry F F θ −− == = °
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72 4 630 630 176.4 604.8 25 25 ⎛⎞ =− = ⎜⎟ ⎝⎠ 1 F j k j k { } 1 176 605 lb Fj k 2 250cos60 250cos135 125 176.777 125 + ° = + Fi j ki j k { } 2 125 177 125 lb + j k Resultant force is { } 125 0.377 480 lb =+= R12 FF F i j k 22 2 (125) ( 0.377) ( 480) 496 lb R F =+ + = Directional angles are 1 125 cos 75.4 496 α == ° 1 0.377 cos 90 496 β ° 1 480 cos 165 496 °
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The length is the magnitude of position vectors So, find , , , and then calculate the magnitudes AD r BD r CD r of each vector
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This homework help was uploaded on 04/04/2008 for the course ME 211 taught by Professor Thouless during the Fall '08 term at University of Michigan.

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ME211solution_1 - (a) Fy sin 45 = - Fn 20 = sin 75 sin 60...

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