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 13.4, Problem 13.173P
To determine

(a)

Boulder will land on the road or beyond if e=0.2

Expert Solution
Check Mark

Answer to Problem 13.173P

The horizontal distance travelled by the rock is 21.31m. Therefore, the rock will land beyond the road.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.173P , additional homework tip  1

α=45°

Rock A falls from a distance of 20m.

h=30md=20m

The total linear momentum of two particles is conserved. Therefore

mAvA+mBvB=mAvA1+mBvB1

The co-efficient of restitution is defined as

vB1vA1=e(vAvB)

For a uniformly accelerated motion

x=x0+v0t+12at2

In above equation

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.173P , additional homework tip  2

Assume v0 as the velocity of rock ‘A’ hits the incline.

v2=u02+2a(xx0)v02=0+2(9.81m/s2)(20m)v0=19.8m/s

Apply conservation of linear momentum in direction t.

mv0sin45°=m(vA1)t

Substitute

(vA1)t=(19.8m/s)sin45°=14m/s

Apply co-efficient of restitution equation in direction n.

0(vA1)n=e(v0cos45°0)

Substitute

(vA1)n=0.2(19.8m/s)cos45°(vA1)n=2.8m/s

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.173P , additional homework tip  3

Velocity (vx)0 in x direction

(vx)0=(vA1)tcos45°+(vA1)nsin45°(vx)0=14cos45°+2.8sin45°=11.879m/s

Velocity in y direction

(vy)0=(vA1)ncos45°(vA1)tsin45°(vy)0=2.8cos45°14sin45°=7.92m/s

Assume t as the time that takes for the rod to reach the ground.

y=y0+v0t+12at2

Substitute

0=307.92t12(9.81)t2

Solve

t=1.794s

Find the horizontal distance xA reached by rock A.

xA=x0+(vx)0t

Substitute

xA=0+(11.879m/s)(1.794s)xA=21.31m

Conclusion:

The horizontal distance travelled by the rock is 21.31m. Therefore, the rock will land beyond the road.

To determine

(b)

Boulder will land on the road or beyond if e=0.1

Expert Solution
Check Mark

Answer to Problem 13.173P

The horizontal distance travelled by the rock is 18.796m. Therefore the rock will land on the road.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.173P , additional homework tip  4

α=45°

Rock A falls from a distance of 20m.

h=30md=20m

The total linear momentum of two particles is conserved. Therefore

mAvA+mBvB=mAvA1+mBvB1

The co-efficient of restitution is defined as

vB1vA1=e(vAvB)

For a uniformly accelerated motion

x=x0+v0t+12at2

In above equation

x - End distance

x0 - Start distance

v0 - Initial velocity

t - Elapsed time

a -Acceleration

Calculation:

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.173P , additional homework tip  5

Assume v0 as the velocity of rock ‘A’ hits the incline.

v2=u02+2a(xx0)v02=0+2(9.81m/s2)(20m)v0=19.8m/s

Apply conservation of linear momentum in direction t.

mv0sin45°=m(vA1)t

Substitute

(vA1)t=(19.8m/s)sin45°=14m/s

Apply co-efficient of restitution equation in direction n.

0(vA1)n=e(v0cos45°0)

Substitute

(vA1)n=0.1(19.8m/s)cos45°(vA1)n=1.4m/s

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.173P , additional homework tip  6

Velocity (vx)0 in x direction

(vx)0=(vA1)tcos45°+(vA1)nsin45°(vx)0=14cos45°+1.4sin45°=10.89m/s

Velocity in y direction

(vy)0=(vA1)ncos45°(vA1)tsin45°(vy)0=1.4cos45°14sin45°=8.91m/s

Assume t as the time that takes for the rod to reach the ground.

y=y0+v0t+12at2

Substitute

0=308.91t12(9.81)t2

Solve

t=1.726s

Find the horizontal distance xA reached by rock A.

xA=x0+(vx)0t

Substitute

xA=0+(10.89m/s)(1.726s)xA=18.796m

Conclusion:

The horizontal distance travelled by the rock is 18.796m. Therefore the rock will land on the road.

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Chapter 13 Solutions

Vector Mechanics For Engineers

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