(a) Find the angular acceleration (magnitude, in rad/s²) of the bridge once it starts to move. 2.22 × Use the rotational analogue of Newton's second law. The drawbridge can be modeled as a rod, with rotation axis about one end. rad/s² (b) How long (in s) does the horse and rider stay in contact with the bridge while it swings downward? 2.88 Think about what happens when the bridge swings downward. Will the acceleration be equal to the acceleration due to gravity? If not, then what does that tell you about the duration of the contact of the horse and the bridge? s (c) Find the angular speed (in rad/s) of the bridge when it strikes the vertical wall below the hinge. 1.27 Apply conservation of mechanical energy, K; + U; = K₁ + Uf. Note that the final kinetic energy will be rotational kinetic energy. A judicious choice of zero level for gravitational potential energy will simplify your calculations. rad/s (d) Find the force (in kN) exerted by the hinge on the bridge immediately after the cable breaks. horizontal component: magnitude direction 12.56 Calculate the tangential acceleration of the center of mass of the bridge, and use your result in Newton's second law for the x- and y-directions. KN to the right vertical component: magnitude 33.35 direction Calculate the tangential acceleration of the center of mass of the bridge, and use your result in Newton's second law for the x- and y-directions. KN upward (e) Find the force (in kN) exerted by the hinge on the bridge immediately before it strikes the wall. horizontal component: magnitude direction vertical component: magnitude 63.98 When the bridge is vertical, is there a horizontal force acting? KN There is no horizontal component. 42.49 ☑ v2 Write Newton's second law for the configuration. Note that the acceleration is centripetal, so a = v² = w²r. Solve for Hy KN direction upward A person on horseback is on a drawbridge which is at an angle = 20.0° above the horizontal, as shown in the figure. The center of mass of the person-horse system is d = 1.35 m from the end of the bridge. The bridge is = 7.00 m long and has a mass of 2,300 kg. A cable is attached to the bridge 5.00 m from the frictionless hinge and to a point on the wall h = 12.0 m above the bridge. The mass of person plus horse is 1,100 kg. Assume the bridge is uniform. Suddenly (and most unfortunately for the horse and rider), the ledge where the bridge usually rests breaks off, and at the same moment the cable snaps and the bridge swings down until it hits the wall.
(a) Find the angular acceleration (magnitude, in rad/s²) of the bridge once it starts to move. 2.22 × Use the rotational analogue of Newton's second law. The drawbridge can be modeled as a rod, with rotation axis about one end. rad/s² (b) How long (in s) does the horse and rider stay in contact with the bridge while it swings downward? 2.88 Think about what happens when the bridge swings downward. Will the acceleration be equal to the acceleration due to gravity? If not, then what does that tell you about the duration of the contact of the horse and the bridge? s (c) Find the angular speed (in rad/s) of the bridge when it strikes the vertical wall below the hinge. 1.27 Apply conservation of mechanical energy, K; + U; = K₁ + Uf. Note that the final kinetic energy will be rotational kinetic energy. A judicious choice of zero level for gravitational potential energy will simplify your calculations. rad/s (d) Find the force (in kN) exerted by the hinge on the bridge immediately after the cable breaks. horizontal component: magnitude direction 12.56 Calculate the tangential acceleration of the center of mass of the bridge, and use your result in Newton's second law for the x- and y-directions. KN to the right vertical component: magnitude 33.35 direction Calculate the tangential acceleration of the center of mass of the bridge, and use your result in Newton's second law for the x- and y-directions. KN upward (e) Find the force (in kN) exerted by the hinge on the bridge immediately before it strikes the wall. horizontal component: magnitude direction vertical component: magnitude 63.98 When the bridge is vertical, is there a horizontal force acting? KN There is no horizontal component. 42.49 ☑ v2 Write Newton's second law for the configuration. Note that the acceleration is centripetal, so a = v² = w²r. Solve for Hy KN direction upward A person on horseback is on a drawbridge which is at an angle = 20.0° above the horizontal, as shown in the figure. The center of mass of the person-horse system is d = 1.35 m from the end of the bridge. The bridge is = 7.00 m long and has a mass of 2,300 kg. A cable is attached to the bridge 5.00 m from the frictionless hinge and to a point on the wall h = 12.0 m above the bridge. The mass of person plus horse is 1,100 kg. Assume the bridge is uniform. Suddenly (and most unfortunately for the horse and rider), the ledge where the bridge usually rests breaks off, and at the same moment the cable snaps and the bridge swings down until it hits the wall.
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter14: Static Equilibrium, Elasticity, And Fracture
Section: Chapter Questions
Problem 31PQ: A wooden door 2.1 m high and 0.90 m wide is hung by two hinges 1.8 m apart. The lower hinge is 15 cm...
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Transcribed Image Text:(a) Find the angular acceleration (magnitude, in rad/s²) of the bridge once it starts to move.
2.22
×
Use the rotational analogue of Newton's second law. The drawbridge can be modeled as a rod, with rotation axis about one end. rad/s²
(b) How long (in s) does the horse and rider stay in contact with the bridge while it swings downward?
2.88
Think about what happens when the bridge swings downward. Will the acceleration be equal to the acceleration due to gravity? If not, then what does that tell you about the duration of the contact of the horse and the bridge? s
(c) Find the angular speed (in rad/s) of the bridge when it strikes the vertical wall below the hinge.
1.27
Apply conservation of mechanical energy, K; + U; = K₁ + Uf. Note that the final kinetic energy will be rotational kinetic energy. A judicious choice of zero level for gravitational potential energy will simplify your calculations. rad/s
(d) Find the force (in kN) exerted by the hinge on the bridge immediately after the cable breaks.
horizontal component:
magnitude
direction
12.56
Calculate the tangential acceleration of the center of mass of the bridge, and use your result in Newton's second law for the x- and y-directions. KN
to the right
vertical component:
magnitude 33.35
direction
Calculate the tangential acceleration of the center of mass of the bridge, and use your result in Newton's second law for the x- and y-directions. KN
upward
(e) Find the force (in kN) exerted by the hinge on the bridge immediately before it strikes the wall.
horizontal component:
magnitude
direction
vertical component:
magnitude
63.98
When the bridge is vertical, is there a horizontal force acting? KN
There is no horizontal component.
42.49
☑
v2
Write Newton's second law for the configuration. Note that the acceleration is centripetal, so a =
v² = w²r. Solve for Hy
KN
direction
upward

Transcribed Image Text:A person on horseback is on a drawbridge which is at an angle = 20.0° above the horizontal, as shown in the figure. The center of mass of the person-horse system is d = 1.35 m from the end of the bridge. The bridge is = 7.00 m long and has a mass of 2,300 kg. A cable is attached to
the bridge 5.00 m from the frictionless hinge and to a point on the wall h = 12.0 m above the bridge. The mass of person plus horse is 1,100 kg. Assume the bridge is uniform. Suddenly (and most unfortunately for the horse and rider), the ledge where the bridge usually rests breaks off,
and at the same moment the cable snaps and the bridge swings down until it hits the wall.
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