a 2.0 kg block rests on a frictionless wedgethat has a 60o incline and an accele

Question

a 2.0 kg block rests on a frictionless wedgethat has a 60o incline and an acceleration a to the right such thatthe mass remains stationary rlative to the wedge (a) draw free bodybody diagram of the block and use it to determine the magnitude ofthe acceleration (b) what would happen if the wedge were given anacceleration larger than this value smaller than this value
I am working this problem (on my own and then following the formulaon cramster) and my values are =55 degrees and the body massis 2kg. I get an acceleration -27.99. I am doing thisproblem on webassign and it assigns random values. It asksfor the magnitude so I enter 27.99 and it appears to bewrong. This acceleartion seems to be awfully high for thissort of problem. I would appreciate any help that isoffered. Thank you for your time. a 2.0 kg block rests on a frictionless wedgethat has a 60o incline and an acceleration a to the right such thatthe mass remains stationary rlative to the wedge (a) draw free bodybody diagram of the block and use it to determine the magnitude ofthe acceleration (b) what would happen if the wedge were given anacceleration larger than this value smaller than this value
I am working this problem (on my own and then following the formulaon cramster) and my values are =55 degrees and the body massis 2kg. I get an acceleration -27.99. I am doing thisproblem on webassign and it assigns random values. It asksfor the magnitude so I enter 27.99 and it appears to bewrong. This acceleartion seems to be awfully high for thissort of problem. I would appreciate any help that isoffered. Thank you for your time.

Explanation / Answer

(a)    consider the free body diagram asshown                                                                      applying the newtons second law of motion weget that    Fn sin30o - w = may    as ay = 0, w = m g    Fn sin30o -m g = 0    Fn sin30o =m g    Fn cos30o =m ax        on dividing we get that    ax / g = cot 30o    ax = g cot30o        = ......... m /s2 (b)    an acceleration of the wedge greaterthan g cot30o woukd require that the normal force exertred on thebody by the    wedge be greater than that given in part(a)    Fn > m g/ sin30o    applying the newtons second law of motion weget that    Fn sin30o - w = may    as ay = 0, w = m g    Fn sin30o -m g = 0    Fn sin30o =m g    Fn cos30o =m ax        on dividing we get that    ax / g = cot 30o    ax = g cot30o        = ......... m /s2 (b)    an acceleration of the wedge greaterthan g cot30o woukd require that the normal force exertred on thebody by the    wedge be greater than that given in part(a)    Fn > m g/ sin30o (b)    an acceleration of the wedge greaterthan g cot30o woukd require that the normal force exertred on thebody by the    wedge be greater than that given in part(a)    Fn > m g/ sin30o

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