you are designing an apparatus to an actor of mass 65 kg who is to \"fly\" down

Question

you are designing an apparatus to an actor of mass 65 kg who is to "fly" down to the stage during a theatre performance. you attach the actor's harness to a 130 kg sandbag with a lightweight steel cable running over two frictionless pulleys (see the picture). you need R=3.0 meters of cable between the harness and the nearest pulley so that the pulley can be hidden behind the stage curtain. for the system to work safely, the sandbag must never left off of the floor as the actor swings. if this happens, the actor could crash into the stage and be injured. let us call the initial angle that the actor's cable makes with the vertical .

you are designing an apparatus to an actor of mass 65 kg who is to ''fly'' down to the stage during a theatre performance. you attach the actor's harness to a 130 kg sandbag with a lightweight steel cable running over two frictionless pulleys (see the picture). you need R=3.0 meters of cable between the harness and the nearest pulley so that the pulley can be hidden behind the stage curtain. for the system to work safely, the sandbag must never left off of the floor as the actor swings. if this happens, the actor could crash into the stage and be injured. let us call the initial angle that the actor's cable makes with the vertical theta. if the actor starts the swing from rest, when he reaches the stage, what is the maximum safe speed that he may be moving at the end of the swing, so that sandbag does not lift off of the floor? (hint: is there a relationship between the maximum safe starting angle theta and the actor's speed at the end of the swing?)

Explanation / Answer

By applying conservation of mechnical energy to the actor and Eath system Kf +Uf = Ki+Ui   .........   (1) Let yi be the initial height of the actor above the floor and vf be his speed at the instant before he lands . 1/2 mactor vf2+0 = 0+mactor gyi   ........ (2) from the geometry of given figure yf = 0 so yi = R-Rcos = R(1-cos) vf2 = 2gR(1-cos)   ............. (3) Applying Newton's second law to the ctor at the bottom of his path T = m actor g +m actor vf2/R      .........   (4) using 3 and 4 mbag g = mactorg + mactor2gR(1-cos)/R cos = 3mactor -m bag / 2mactor          = 3*65 kg - 130 kg /2*65 kg          = 0.50        = 600 Now from the equation 3 the maximum safe speed (when =900) of actor is vf2 = 2gR(1-cos)      = 2*(9.8 m/sec2)(3.0 m) (1-cos900)      = 58.8 m2/sec2 vf = 7.66 m/sec so the maximum safe speed of actor is vf = 7.66 m/sec          = 3*65 kg - 130 kg /2*65 kg          = 0.50        = 600 Now from the equation 3 the maximum safe speed (when =900) of actor is vf2 = 2gR(1-cos)      = 2*(9.8 m/sec2)(3.0 m) (1-cos900)      = 58.8 m2/sec2 vf = 7.66 m/sec so the maximum safe speed of actor is vf = 7.66 m/sec Now from the equation 3 the maximum safe speed (when =900) of actor is vf2 = 2gR(1-cos)      = 2*(9.8 m/sec2)(3.0 m) (1-cos900)      = 58.8 m2/sec2 vf = 7.66 m/sec vf = 7.66 m/sec so the maximum safe speed of actor is vf = 7.66 m/sec vf = 7.66 m/sec

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