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Unformatted text preview: alexander (jra2623) – homework 24 – Turner – (92510) 1 This printout should have 11 questions. Multiplechoice 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 θ . W P F θ R Determine the torque equation about the point P. 1. mg R = F (1 + cos θ ) R 2. mg R = F (1 sin θ ) R 3. mg R = F (1 cos θ ) R 4. mg R = F (1 + sin θ ) R correct 5. mg R = 2 F R 6. mg R = F R Explanation: W P F θ R R ( 1 + s i n θ ) A θ The clockwise torque about point P is τ CW = AP F = R (1 + sin θ ) F and the counterclockwise torque is τ CCW = mg R . From rotational equilibrium, τ CCW = τ CW mg R = R (1 + sin θ ) 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 α β 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 β sin( α + β ) 2. F A = M g sin α cos( α + β ) 3. F A = M g cos β sin( α + β ) correct 4. F A = M g cos β cos( α + β ) 5. F B = M g cos α cos( α + β ) 6. F A = M g sin β sin( α + β ) Explanation: Mg A B F A F A cos α F B F B cos β F A sin α F B sin β α β At equilibrium, summationdisplay vector F = 0 summationdisplay vector τ = 0 . Call the normal forces F A and F B . They make angles α and β with the vertical, so applying translational rotation horizontally summationdisplay F x = 0 F A cos α F B cos β = 0 F B = F A cos α cos β , and vertically summationdisplay F y = 0 F A sin α M g + F B sin β = 0 F A sin α + F A sin β cos α cos...
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 Summer '08
 Kaplunovsky
 Force, Mass, Work, Orders of magnitude, Wire

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