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Hi.
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So in our question we have michael jordan that has a portable basketball hoop.
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Now he is set to basket at 8 feet high so that he can practice dunking and he slams the ball right through the hoop and then hangs onto it 1 .1 meters in front of the edge of the portable hoop's base.
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Now the mass of the whole portable basketball hoop is 70 kg.
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The center of gravity of the portable hoop is 1 meter behind the front edge of the base.
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Now we need to determine the torque that is produced around the front of the hoop base by the 600 newton force that jordan exerts on the front of the rim.
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Now let the end of the rim be point the center of gravity be hoop b and the front edge of the rim be point o.
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First we'll determine the talk about point o which is due to the 6 .5.
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Newton force applied by jordan now we know that torque is equivalent to force into the perpendicular distance that is covered in here simply applying the value of force being 600 newton and the perpendicular distance of 1 .1 meter we get the talk about 0 .0 to be equal to 660 newton meter next we'll determine the torque required to tip the hoop now for this we know that the hoop is balanced by the self -weight acting through the center of gravity at point b.
01:59
Now the torque due to the self -weight will be equal to 70 into 9 .8 meter per second square multiplied by 1 meter...