VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter 11.4, Problem 11.127P

Coal discharged from a dump truck with an initial velocity (vc)0 = 5 ft/s ⦫ 50° falls onto conveyor belt B. Determine the required velocity vB of the belt if the relative velocity with which the coal hits the belt is to be (a) vertical, (b) as small as possible.

Fig. P11.127

Chapter 11.4, Problem 11.127P, Coal discharged from a dump truck with an initial velocity (vc)0 = 5 ft/s  50 falls onto conveyor

(a)

Expert Solution
Check Mark
To determine

The required velocity (vB) of the belt if the relative velocity with which the coal hit the belt is to be vertical.

Answer to Problem 11.127P

The required velocity (vB) of the belt if the relative velocity with which the coal hit the belt is to be vertical is 3.26ft/s_ at an angle of 10°_.

Explanation of Solution

Given Information:

The initial (vC)0 velocity of coal 5ft/s at an angle θ of 50°.

The angle (α) between the horizontal surface and the belt is 10°.

Calculation:

Calculate the x component initial velocity ((vC)0)x of coal using the relation:

((vC)0)x=(vC)0cosθ

Substitute 5ft/s for (vC)0 and 50° for θ

((vC)0)x=5cos(50°)=3.2139ft/s

Calculate the x component velocity (vC)x of coal using the relation:

(vC)x=((vC)0)x(vC)x=3.2139ft/s

Calculate the y component initial velocity ((vC)0)y of coal using the relation:

((vC)0)y=(vC)0sinθ

Substitute 5ft/s for (vC)0 and 50° for θ.

((vC)0)y=5sin(50°)=3.830ft/s

Calculate the y component kinematic velocity (vC)y of coal using the relation:

(vC)y=((vC)0)y2+2ayd

Here, d is distance between the dump truck and belt.

Substitute 3.830ft/s for ((vC)0)y and g for ay.

(vC)y2=3.8302+2(g)d(vC)y2=3.83022gd

Substitute 32.2ft/s2 for g and 4ft for d.

(vC)y2=3.83022(32.2)(4)(vC)y=272.2689(vC)y=16.501ft/s

Calculate the angle (β) between the velocity C and horizontal line.

tanβ=(vC)y(vC)x

Substitute 16.501ft/s for (vC)y and 3.2139ft/s for (vC)x.

tanβ=16.5013.2139=tan1(5.134)=78.99°

Calculate the velocity (vC) of coal using the relation:

vC=(vC)x2+(vC)y2

Substitute 3.2139ft/s for (vC)x and 16.501ft/s for (vC)y.

vC=(3.2139)2+(16.501)2=10.329+272.28=16.81ft/s

Write the expression for velocity vector (vC) of the coal:

vC=(3.2139)i+(16.501)j

Write the expression for velocity vector (vB) of the belt:

vB=vB(cosαi+sinαj)

Substitute 10° for α.

vB=vB(cos(10°)i+sin(10°)j)=0.985vBi+0.1736vBj

Write the expression for relative velocity vector (vC/B):

vC/B=vC-vB (1)

When the belt positioned vertical, then x component of relative velocity (vC/B)x become zero.

Write the y component relative velocity vector equation (vC/B)y as below:

Substitute (3.2139)i+(16.501)j for vC and 0.985vBi+0.1736vBj for vB in Equation (1).

(vC/B)yj=(3.2139)i+(16.501)j(0.985vBi+0.1736vBj)

Write the y component relative velocity vector equation (vC/B)y as below:

Substitute (3.2139)i+(16.501)j for vC and 0.985vBi+0.1736vBj for vB in Equation (1).

(vC/B)yj=(3.2139)i+(16.501)j(0.985vBi+0.1736vBj)

Write the x component relative velocity vector equation (vC/B)x:

Substitute 0 for (vC/B)x, (3.2139)i+(16.501)j for vC and 0.985vBi+0.1736vBj for vB in Equation (1).

0i=(3.2139)i+(16.501)j(0.985vBi+0.1736vBj)0i=(3.2139)i+0.985vBi

Solve the above equation for velocity (vB) of block B.

vB=3.21390.985=3.26ft/s

Therefore, the required velocity (vB) of the belt if the relative velocity with which the coal hit the belt is to be vertical is 3.26ft/s_ at an angle of 10°_.

(b)

Expert Solution
Check Mark
To determine

The required velocity (vB) of the belt if the relative velocity with which the coal hit the belt is to be as small as possible.

Answer to Problem 11.127P

The required velocity (vB) of the belt if the relative velocity with which the coal hit the belt is to be as small as possible is 0.3ft/s_ at an angle of 10°_.

Explanation of Solution

Given Information:

The initial (vC)0 velocity of coal 5ft/s at an angle θ of 50°.

The angle (α) between the horizontal surface and the belt is 10°.

Calculation:

Write vector for relative velocity (vC/B) of C with respect to B:

Substitute (3.2139)i+(16.501)j for vC and 0.985vBi+0.1736vBj for vB in Equation (1).

(vC/B)xi+(vC/B)yj=(3.2139)i+(16.501)j(0.985vBi+0.1736vBj)

Separate the i and j component.

Consider i component.

(vC/B)xi=(3.2139)i+0.985vBi

Consider j component.

(vC/B)yj=(16.501)j(0.1736vBj)

Find the expression for (vC/B)2 using the relation:

(vC/B)2=(vC/B)x2+(vC/B)y2

Substitute (3.2139)i+0.985vBi for (vC/B)x and (16.501)j(0.1736vBj) for (vC/B)y.

(vC/B)2=((3.2139)+0.985vB)2+((16.501)(0.1736vB))2={10.329+0.970vB2+2(3.2139×0.985vB)+272.283+0.0301vB2 2(16.501×0.1736vB)}=10.329+0.970vB26.331vB+272.283+0.0301vB2+5.729vB=282.6120.602vB+vB2 (2)

Calculate the required velocity (vB) of the belt if the relative velocity with which the coal hit the belt is to be as small as possible:

Differentiate the equation 2 with respective to velocity (vB)

d(vC/B)2dvB=282.6120.602vB+vB2=0.602+2vB

Solve the above equation for (vB):

vB=0.6022=0.3ft/s

Therefore, the required velocity (vB) of the belt if the relative velocity with which the coal hit the belt is to be as small as possible is 0.3ft/s_ at an angle of 10°_.

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Chapter 11 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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