Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 17.3, Problem 17.118P

A uniformly loaded square crate is released from rest with its comer D directly above A; it rotates about A until its comer B strikes the floor, and then rotates about B. The floor is sufficiently rough to prevent slipping and the impact at B is perfectly plastic. Denoting by ω 0 the angular velocity of the crate immediately before B strikes the floor, determine (a) the angular velocity of the crate immediately after B strikes the floor, (b) the fraction of the kinetic energy ofthe crate lost during the impact, (c) the angle θ through which the crate will rotate after B strikes the floor.

  Chapter 17.3, Problem 17.118P, A uniformly loaded square crate is released from rest with its comer D directly above A; it rotates

Expert Solution
Check Mark
To determine

(a)

The angular velocity of the crate immediately after B strikes the floor.

Vector Mechanics For Engineers, Chapter 17.3, Problem 17.118P , additional homework tip  1

Answer to Problem 17.118P

The angular velocity of the square box after striking B on the floor will be 14 th of angular velocity before B strikes on the floor.

ω=1/4ω0

Explanation of Solution

Given:

Square box which is uniformly loaded is released on the floor when its corner D is directly above the A. square box tends to rotate till it strikes the floor. The floor is made anti-slipping and impact at B is perfectly plastic.

Concept:

According to impulse momentum principle,

Vector Mechanics For Engineers, Chapter 17.3, Problem 17.118P , additional homework tip  2

Calculation:

Let’s consider,

M = mass of a square box.

C = length of the side.

Moment of inertia = I=1/2m(c2+c2)=1/6mc2

We can say,

v0=rG/Axω0=1/2ω0v=rG/Bxω=1/2ω

Taking moment at A,

Iω0+0=Iω+rG/Bx mv1/6mc2ω0=1/6mc2ω+(1/2)m(1/2)1/6mc2ω0=1/6mc2ω+[/4]1/6mc2ω0=1/6mc2ω+2c2mω/4mc2ω0/6=mc2ω/6+2c2mω/4mc2ω0/6=mc2ω/2(1/3+2/2)mc2ω0/6=4/1(mc2ω/2)3/4x2/6xmc2ω0/mc2=ω1/4ω0=ω

Conclusion:

The angular velocity of the square box after striking B on the floor will be 14 of the angular velocity before B strikes on the floor.

Expert Solution
Check Mark
To determine

(b)

The fraction of kinetic energy lost during the impact.

Answer to Problem 17.118P

The fractions of energy lost during impact conditions are.

Explanation of Solution

Given:

Square box which is uniformly loaded is released on the floor when its corner D is directly above the A. square box tends to rotate till it strikes the floor. The floor is made anti-slipping and impact at B is perfectly plastic.

Calculation:

Kinetic energy before impact,

E1=1/2 Iω02+1/2mv02=1/2(1/6mc2)ω02+1/2m(1/2cω0)2=mc2ω02/12+m/2(2/4c2ω02)=mc2ω02/12+mc2ω02/4=mc2ω02(1/12+1/4)E1=1/3 mc2ω02

Similarly kinetic energy after impact,

E2=1/2 Iω2+1/2mv2=1/2(1/6mc2)ω2+1/2m(1/2cm)2=mc2ω2/12+mc2ω2/4=1/3mc2ω2=1/3mc2(1/4ω0)2E2=1/48mc2ω02

Combining E1 and E2 to find fraction of energy cost,

E1E2/E1=1/3mc2ω0 1/48mc2ω0/1/3mc2ω0=(1/31/48)mc2ω02/(1/3)mc2ω02=1/31/48/1/3=(1/3/(1/3)(1/48)/(1/3)=3/33/48=11/1615/16

Conclusion:

The fraction of energy lost during impact conditions is 1516.

Expert Solution
Check Mark
To determine

(c)

The angle through which the crater will rotate after B strikes the floor.

Answer to Problem 17.118P

Angle made by corner A of square box and floor will be 1.500

Explanation of Solution

Given:

Square box which is uniformly loaded is released on the floor when its corner D is directly above the A. square box tends to rotate till it strikes the floor. The floor is made anti-slipping and impact at B is perfectly plastic.

Concept:

As per law of conservation of energy before impact,

T0+V0=T1+V1

And that of after impact,

T3+V3=T2+V2T0=0,T3=0, T2=1/16Tv0=mg(1/22c)v1=v2=mg(1/22c)v3=mgh3

Before impact,

T0+V0=T1+V1T1=V0V1=mg(1/22c)=mg(1/22c)mg(1/2c)=mg2c/2mgc/2T1=mgc/2(21)

After impact,

T3+V3=T2+V2T2=V3V2=mgh3mgc/2T2=mg(h3c/2)

As,

T2=1/16T11/16mgc/2(21)=mg(h3c/2)21/32c=h3c/2h3=((21)c/32+c/2=((21)c+16c/32h3=((21)+16)c/32

But

Vector Mechanics For Engineers, Chapter 17.3, Problem 17.118P , additional homework tip  3

Therefore, from geometry,

h3=1/22csin(θ+450)

Equating both equations of h3

[(21)+16/32]c=2c/2sin(θ+450)[(21)+16/32]c×2/2c=sin(θ+450)21/16+1=sin(θ+450)1.0258= sin(θ+450)θ+45=46.50θ=46.5045=1.50

Conclusion:

1.500  the angle made by corner A of square box and floor.

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Chapter 17 Solutions

Vector Mechanics For Engineers

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