Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 20P

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 angular acceleration of the bridge just as it starts to move. (c) Find the angular speed of the bridge when it strikes the wall below the hinge. Find the force exerted by the hinge on the bridge (d) immediately after the cable breaks and (e) immediately before it strikes the castle wall.

Chapter 12, Problem 20P, Review. While Lost-a-Lot ponders his next move in the situation described in Problem 11 and

(a)

Expert Solution
Check Mark
To determine

The time that Lost a Lot stay in contact with the bridge while it swings downward.

Answer to Problem 20P

The time that Lost a Lot stay in contact with the bridge while it swings downward is 0s .

Explanation of Solution

Given info: The length of the uniform bridge is 8.00m , the mass of the bridge is 2000kg , the height of the castle wall is 12.0m and the angle made by the bridge from of horizontal is 20.0° .

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 0s .

(b)

Expert Solution
Check Mark
To determine

The angular acceleration of the bridge just as it starts to move.

Answer to Problem 20P

The angular acceleration of the bridge just as it starts to move is 1.73rad/s2 .

Explanation of Solution

Given info: The length of the uniform bridge is 8.00m , the mass of the bridge is 2000kg , the height of the castle wall is 12.0m and the angle made by the bridge from of horizontal is 20.0° .

The mass moment of inertial along the centroid is,

I=13ml2

The total moment along the centroid is,

M=mg(l2)cosθ

Here,

I is the mass moment o0f inertia along the centroid.

m is the mass of the bridge.

l is the length of the bridge.

g is the acceleration due to gravity.

θ is the angle made by the bridge from of horizontal.

Total moment along the centroid is,

M=Iα (1)

Here,

M is the moment along centroid of bridge.

α is the angular acceleration.

Substitute mg(l2)cosθ for M and 13ml2 for I in equation (1).

mg(l2)cosθ=13ml2αα=3gcosθ2l (2)

Substitute 9.81m/s2 for g , 20.0° for θ and 8m for l in equation (2)

α=3(9.81m/s2)cos(20.0°)2(8.00m)=1.73rad/s2

Conclusion:

Therefore, the angular acceleration of the bridge just as it starts to move is 1.73rad/s2 .

(c)

Expert Solution
Check Mark
To determine

The angular speed of the bridge when it strikes the wall below the hinge.

Answer to Problem 20P

The angular speed of the bridge when it strikes the wall below the hinge is 2.22rad/s .

Explanation of Solution

Given info: The length of the uniform bridge is 8.00m , the mass of the bridge is 2000kg , the height of the castle wall is 12.0m and the angle made by the bridge from of horizontal is 20.0° .

The total height of the wall from the point of hinge is,

h=l2+l2sinθ=l2(1+sinθ)

Here,

h is the height of the castle wall.

From the conservation of energy, the total potential energy will be equal to the rotational energy is,

mgh=12Iω2 (2)

Here,

ω is the angular speed.

Substitute l2(1+sinθ) for h and 13ml2 for I in equation (2) to find ω .

mgl2(1+sinθ)=12(13ml2)ω2ω=3g(1+sinθ)l (3)

Substitute 9.81m/s2 for g , 20.0° for θ and 8m for l in equation (3).

ω=3(9.81m/s2)(1+sin(20.0°))8m=2.22rad/s

Conclusion:

Therefore, the angular speed of the bridge when it strikes the wall below the hinge is 2.22rad/s .

(d)

Expert Solution
Check Mark
To determine

The force exerted by the hinge on the bridge immediately after the cable breaks.

Answer to Problem 20P

The force exerted by the hinge on the bridge immediately after the cable breaks is (4.72i^+6.62j^)kN .

Explanation of Solution

Given info: The length of the uniform bridge is 8.00m , the mass of the bridge is 2000kg , the height of the castle wall is 12.0m and the angle made by the bridge from of horizontal is 20.0° .

From Figure (1), the tangential acceleration is,

at=l2α

Here,

at is the tangential acceleration.

Substitute 8.00m for l and 1.73rad/s2 for α in above equation to find at .

at=8.00m2(1.73rad/s2)=6.92m/s2

Thus, the tangential acceleration is 6.92m/s2 .

The acceleration along the horizontal is,

ax=atsinθ

Here,

ax is the acceleration along the horizontal.

The acceleration along the vertical is,

ay=(gatcosθ)

Here,

ay is the acceleration along the vertical

Force along the horizontal direction is,

Fx=max (4)

Here,

Fx is the force along the horizontal.

Substitute atsin20° for ax in equation (4).

Fx=matsinθ (5)

Substitute 2000kg for m , 20.0° for θ and 6.92m/s2 for at in equation (5) to find Fx .

Fx=(2000kg)(6.92m/s2)sin(20°)=4.72×103N(103kN1N)=4.72kN

Thus, the force along the horizontal is 4.72kN .

Force along the vertical is,

Fy=may (6)

Here,

Fy is the force along the vertical.

Substitute (gatcosθ) foe ay in equation (6).

Fy=m(gatcosθ) (7)

Substitute 2000kg for m , 20.0° for θ , 9.81m/s2 for g and 6.92m/s2 for at in equation (7) to find Fy .

Fy=(2000kg)(9.81m/s2(6.92m/s2)cos20.0°)=6.62N(103kN1N)=6.62kN

Thus, the vertical force is 6.62kN .

The force exerted by the hinge on the bridge is,

F=Fx+Fy (8)

Here,

F is the force exerted by the hinge on the bridge.

Substitute 6.62kN for Fy and 4.72kN for Fx in equation (8) to find F .

F=(4.72i^+6.62j^)kN

Conclusion:

Therefore, the force exerted by the hinge on the bridge immediately after the cable breaks is (4.72i^+6.62j^)kN .

(e)

Expert Solution
Check Mark
To determine

The force exerted by the hinge on the bridge immediately before strikes the cable wall.

Answer to Problem 20P

The force exerted by the hinge on the bridge immediately before strikes the cable wall is 59.1kN .

Explanation of Solution

Given info: The length of the uniform bridge is 8.00m , the mass of the bridge is 2000kg , the height of the castle wall is 12.0m and the angle made by the bridge from of horizontal is 20.0° .

The acceleration along the vertical is,

ay=ω2l2

From Newton’s second law, the total force along the vertical is,

Fy=mayFymg=mayFy=m(g+ay) (9)

Substitute ω2l2 for ay in equation (9) to find Fy .

Fy=m(g+ω2l2) (10)

Substitute 2000kg for m , 2.22rad/s for ω , 9.81m/s2 for g and 8.00m for l in equation (10) to find Fy .

Fy=(2000kg)(9.81m/s2+(2.22rad/s)28.00m2)=59.1N(103kN1N)=59.1kN

Conclusion:

Therefore, the force exerted by the hinge on the bridge immediately before strikes the cable wall is 59.1kN .

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Chapter 12 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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