hw24 - alexander (jra2623) homework 24 Turner (92510) 1...

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Unformatted text preview: alexander (jra2623) homework 24 Turner (92510) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A solid sphere of radius R and mass m is held against a wall by a string being pulled at an angle . W P F R Determine the torque equation about the point P. 1. mg R = F (1 + cos ) R 2. mg R = F (1- sin ) R 3. mg R = F (1- cos ) R 4. mg R = F (1 + sin ) R correct 5. mg R = 2 F R 6. mg R = F R Explanation: W P F R R ( 1 + s i n ) A The clockwise torque about point P is CW = AP F = R (1 + sin ) F and the counterclockwise torque is CCW = mg R . From rotational equilibrium, CCW = CW mg R = R (1 + sin ) F . 002 (part 1 of 2) 10.0 points A solid sphere of radius R and mass M is placed in a wedge as shown in the figure. The inner surfaces of the wedge are frictionless. M R A B alexander (jra2623) homework 24 Turner (92510) 2 Determine the force exerted by the wedge on the sphere at the left contact point. 1. F A = 2 M g sin sin( + ) 2. F A = M g sin cos( + ) 3. F A = M g cos sin( + ) correct 4. F A = M g cos cos( + ) 5. F B = M g cos cos( + ) 6. F A = M g sin sin( + ) Explanation: Mg A B F A F A cos F B F B cos F A sin F B sin At equilibrium, summationdisplay vector F = 0 summationdisplay vector = 0 . Call the normal forces F A and F B . They make angles and with the vertical, so applying translational rotation horizontally summationdisplay F x = 0 F A cos - F B cos = 0 F B = F A cos cos , and vertically summationdisplay F y = 0 F A sin - M g + F B sin = 0 F A sin + F A sin cos cos...
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hw24 - alexander (jra2623) homework 24 Turner (92510) 1...

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