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 14.3, Problem 14.68P

Coal is being discharged from a first conveyor belt at the rate of 120 kg/s. It is received at A by a second belt discharges it again at B. Knowing that v 1 = 3 m/s and v 2 = 4.25 m/s and that the second belt assembly and the coal it supports have a total mass of 472 kg, determine the components of the reactions at C and D.

    Chapter 14.3, Problem 14.68P, Coal is being discharged from a first conveyor belt at the rate of 120 kg/s. It is received at A by

Expert Solution & Answer
Check Mark
To determine

The components of reaction at C and D.

Answer to Problem 14.68P

The y component of reaction at D is 2900N and the x component of reaction at D is 0N, the y component of reaction at C is 2362.7N and the x component of reaction at C is 90N.

Explanation of Solution

Given:

Rate of discharge of coal is 120 kg/s.

Velocity v1 is 3 m/s.

Velocity v2 is 4.25 m/s.

Total mass is 472kg.

Concept used:

Draw the diagram to illustrate the path travelled by coal on the belt as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.3, Problem 14.68P , additional homework tip  1

The path followed by the coal over belt can illustrated as:

Vector Mechanics for Engineers: Dynamics, Chapter 14.3, Problem 14.68P , additional homework tip  2

Write the expression for the velocity component in y direction.

(v1)y=2gh ...... (1)

Here, (v1)y is the velocity component in y direction, g is acceleration due to gravity and h is the height from which coal fall.

Write the expression for angle θ in triangle EBF.

θ=tan1(BFEF) ...... (2)

Here, θ is the angle of inclination of conveyor.

Write the expression for the moment about point C.

[(Δm)(v1)x(1.2m)+(Δm)(v1)y(0.75m)+(W)1.8(D)(3m)]=[(Δm)(v2cosθ)(2.4 m)(Δm)(v2sinθ)(3 m)]

Simplify the above expression.

(D)(3m)=[1.2(Δm)(v1)x+0.75(Δm)(v1)y+1.8(W)2.4(Δm)(v2cosθ)+3(Δm)(v2sinθ)] ...... (3)

Here, D is the reaction at point D, Δm is the mass of coal entering and leaving the system, W is the rate of discharge of coal, (v1)x is the velocity component in x direction and v2 is velocity.

Write the expression for the x component of reaction at C.

CxΔt=(Δm)Δtv2cosθΔmΔt(v1)x ...... (4)

Here, Cx is the x component of reaction at C.

Write the expression for y component of C.

CyΔt=Δmv2sinθΔt+Δm(v1)yΔtDΔt+WΔt ...... (5)

Here, Cy is the y component of reaction at C.

Calculation:

Substitute 9.81 m/s2 for g and 0.545 m for h in equation (1).

(v1)y=2(9.80)(0.545)=3.27m/s

Substitute 1.2 for BF and 2.25 for EF in equation (2).

θ=tan1(1.22.25)θ=28.07°

Substitute 472kg for W, 120 kg for Δm, 28.07° for θ, 3.27 m/s for (v1)y, 3 for (v1)x and 4.25 m/s for v2 in equation (3).

3D=[(472)(9.81)(1.8)+120(3)(1.2)+120(3.27)(0.75)120(4.25)(2.4)(cos28.07°)+120(4.25)(3)(sin28.07°)]3D=8700D=2900 N

Substitute 120kg/s for Δm, 28.07° for θ, 4.25 m/s for v2 and 3 m/s for (v1)x in equation (4).

CxΔt=120(4.25)Δtcos28.07°120Δt(3)Cx=90 N

Substitute 120kg/s for Δm, 28.07° for θ, 4.25 m/s for v2, 472kg for W and 2900N for D in equation (5).

CyΔt=120Δt(4.25)sin28.07°+120Δt(3.27)2900Δt+472Δt(9.81)Cy=239.98+392.42900+4630.32=2362.7N

The y component of reaction at D is 2900N and the x component of reaction at D is 0N, the y component of reaction at C is 2362.7N and the x component of reaction at C is 90N.

Conclusion:

Thus, the y component of reaction at D is 2900N and the x component of reaction at D is 0N, the y component of reaction at C is 2362.7N and the x component of reaction at C is 90N.

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Chapter 14 Solutions

Vector Mechanics for Engineers: Dynamics

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