Concept explainers
Review. While Lost-a-Lot ponders his next move in the situation described in Problem 11 and illustrated in Figure P12.11, the enemy attacks! An incoming projectile breaks off the stone ledge so that the end of the drawbridge can be lowered past the wall where it usually rests. In addition, a fragment of the projectile bounces up and cuts the drawbridge cable! The hinge between the castle wall and the bridge is frictionless, and the bridge swings down freely until it is vertical and smacks into the vertical castle wall below the castle entrance. (a) How long does Lost-a-Lot stay in contact with the bridge while it swings downward? (b) Find the
(a)
The time that Lost a Lot stay in contact with the bridge while it swings downward.
Answer to Problem 12.20P
The time that Lost a Lot stay in contact with the bridge while it swings downward is
Explanation of Solution
The length of the uniform bridge is
There is no time interval as the Lost a Lot stay in contact with the bridge while it swings downward because the horse feet loose contact with the down bridge as soon as it begins to move because the vertical acceleration act on the feet is greater than the acceleration due to gravity due to that the horse is in the air and moves upward with a vertical component of acceleration.
Conclusion:
Therefore, the time that Lost a Lot stay in contact with the bridge while it swings downward is
(b)
The angular acceleration of the bridge just as it starts to move.
Answer to Problem 12.20P
The angular acceleration of the bridge just as it starts to move is
Explanation of Solution
The mass moment of inertial along the centroid is,
The total moment along the centroid is,
Here,
Total moment along the centroid is,
Here,
Substitute
Substitute
Conclusion:
Therefore, the angular acceleration of the bridge just as it starts to move is
(c)
The angular speed of the bridge when it strikes the wall below the hinge.
Answer to Problem 12.20P
The angular speed of the bridge when it strikes the wall below the hinge is
Explanation of Solution
The total height of the wall from the point of hinge is,
Here,
From the conservation of energy, the total potential energy will be equal to the rotational energy is,
Here,
Substitute
Substitute
Conclusion:
Therefore, the angular speed of the bridge when it strikes the wall below the hinge is
(d)
The force exerted by the hinge on the bridge immediately after the cable breaks.
Answer to Problem 12.20P
The force exerted by the hinge on the bridge immediately after the cable breaks is
Explanation of Solution
From Figure (1), the tangential acceleration is,
Here,
Substitute
Thus, the tangential acceleration is
The acceleration along the horizontal is,
Here,
The acceleration along the vertical is,
Here,
Force along the horizontal direction is,
Here,
Substitute
Substitute
Thus, the force along the horizontal is
Force along the vertical is,
Here,
Substitute
Substitute
Thus, the vertical force is
The force exerted by the hinge on the bridge is,
Here,
Substitute
Conclusion:
Therefore, the force exerted by the hinge on the bridge immediately after the cable breaks is
(e)
The force exerted by the hinge on the bridge immediately before strikes the cable wall.
Answer to Problem 12.20P
The force exerted by the hinge on the bridge immediately before strikes the cable wall is
Explanation of Solution
The acceleration along the vertical is,
From Newton’s second law, the total force along the vertical is,
Substitute
Substitute
Conclusion:
Therefore, the force exerted by the hinge on the bridge immediately before strikes the cable wall is
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Chapter 12 Solutions
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