Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
11th Edition
ISBN: 9780077687342
Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 13.1, Problem 13.12P

Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to conveyor belt which moves with a velocity of 2 m/s. Knowing that d = 7.5 m and μ k = 0.25 between the packages and all surfaces, determine (a) the speed of the package at C, (b) the distance a package will slide on the conveyor belt before it comes to rest relative to the belt.

Expert Solution & Answer
Check Mark
To determine

(a)

The speed of package at point C.

Answer to Problem 13.12P

The speed of package at point C is 2.9 m/s.

Explanation of Solution

Given:

Velocity of packages at point A is 1 m/s.

Velocity of conveyor is 2 m/s.

Coefficient of kinetic friction is 0.25.

Angle of inclination of conveyor is 30°.

Concept used:

Refer to Fig P13.12; consider the movement of package from point A to point B.

Draw FBD of block at point A.

Vector Mechanics for Engineers: Dynamics, Chapter 13.1, Problem 13.12P , additional homework tip  1

Write the expression for the kinetic energy of the package at point A.

TA=12mvA2   ...... (1)

Here, TA is the kinetic energy of the package at point A, m is the mass of the package and vA is the initial velocity of the package.

Write the expression for the kinetic energy of the package at point B.

TB=12mvB2   ...... (2)

Here, TB is the kinetic energy of the package at point B and vB is the velocity of the package at point B.

Write the expression for the work done for moving package from point A to point B.

UAB=(mgsinθF)(d)   ...... (3)

Here, UAB is the work done to move package from point A to point B, g is the acceleration due to gravity, θ is the angle of inclination, d is the distance moved by the package from A to B and F is the frictional force acting on the package.

For the system to be in static equilibrium the summation of all the forces in the direction of motion should be zero.

F=0

Take summation of forces in the direction of motion.

F=Nmgcosθ

Substitute 0 for F in the above expression and rearrange for N.

N=mgcosθ

Write the expression for the force of friction acting on the package.

F=μkN

Substitute mgcosθ for N in the above expression.

F=μk(mgcosθ)

Substitute μk(mgcosθ) for F in the equation (3).

UAB=(mgsinθμk(mgcosθ))(d)   ...... (4)

Write the expression as per work energy principle for package.

TA+UAB=TB   ...... (5)

Draw the FBD of the package at point B ,

Vector Mechanics for Engineers: Dynamics, Chapter 13.1, Problem 13.12P , additional homework tip  2

Write the expression for the kinetic energy at point C.

TC=12mvC2   ...... (6)

Here, TC is the kinetic energy of package at point C and vC is the velocity of package at point C.

Write the expression for the work done for moving package from point B to point C.

UBC=μkmgd'   ...... (7)

Here, UBC is the work done for moving the package from B to C, d' is the distance between the points B and C.

Write the expression as per work energy principle for package.

TB+UBC=TC   ...... (8)

Calculation:

Substitute 1 m/s for vA is equation (1).

TA=12m(1)2=0.5m

Substitute 9.81 m/s2 for g, 30° for θ, 7.5 m for d and 0.25 for μk in equation (4).

UAB=(m(9.81)sin30°(0.252.781)(m(9.81)cos30°))(7.5)=20.858m

Substitute 12mvB2 for TB, 0.5m for TA and 20.858m for UAB in equation (5).

0.5m+20.858m=12mvB2(0.5+20.858)m=12mvB2

Rearrange the above expression for vB.

vB=42.716=6.535 m/s

Substitute 0.25 for μk, 9.81 m/s2 for g and 7 m for d' in equation (7).

UBC=(0.25)m(9.81)(7)=17.168m

Substitute 12mvB2 for TB, 12mvC2 for TC and 17.168m for UBC in equation (8).

12mvB217.168m=12mvC20.5vB217.168=0.5vC2

Substitute 6.535 m/s for vB in above expression.

0.5(6.535)217.168=0.5vC221.35317.168=0.5vC2

Simplify the above expression for vC.

vC=8.37=2.893 m/s2.9m/s

The speed of package at point C is 2.9 m/s.

Conclusion:

Thus, the speed of package at point C is 2.9 m/s.

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Chapter 13 Solutions

Vector Mechanics for Engineers: Dynamics

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