VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
12th Edition
ISBN: 9781260265521
Author: BEER
Publisher: MCG
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Chapter 16.2, Problem 16.89P
To determine

The component of reaction at point C at θ=0°.

Expert Solution & Answer
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Answer to Problem 16.89P

The reaction at point C is 30.9918 lb in the direction of 83.18° measured form positive x -axis.

Explanation of Solution

Given:

Weight of rod AB is 2lb.

Weight of rod BC is 4lb.

Angular velocity is 10rad/s.

Angle θ is 0°.

Concept used:

Write the expression for the moment of inertia of rod AB.

IAB=WABLAB212g ...... (1)

Here, IAB is the moment of inertia of rod AB, g is acceleration due to gravity, WAB is the weight of bar AB and LAB is the length of bar AB.

Write the expression for the moment of inertia of rod BC.

IBC=WBCLBC212g ...... (2)

Here, IBC is the moment of inertia of rod BC, WBC is the weight of bar BC and LBC is the length of bar BC.

Draw the diagram to illustrate the geometry of system as shown below:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 16.2, Problem 16.89P , additional homework tip  1

From above figure calculate the distance between Point C and D,

rCD=LBC2+LBD2 ...... (3)

Here, rCD is the distance between C and D and LBD is the distance between point B and D.

Write the expression for distance between B and E.

rBE=LBC2 ...... (4)

Here, rBE is the distance between points B and E.

Write the expression for angle β.

tanβ=LBDLBC ...... (5)

Write the expression for tangential acceleration of rod AB.

(aAB)t=αrCD

Here, (aAB)t is the tangential component of acceleration of rod AB and α is the angular acceleration of rod.

Write the expression for centripetal acceleration of rod AB.

(aAB)n=ω2rCD

Here, (aAB)n is the centripetal acceleration and ω is the angular velocity of rod.

Write the expression for tangential acceleration of rod BC.

(aBC)t=αrCE

Here, (aBC)t is the tangential component of acceleration of rod BC.

Write the expression for centripetal acceleration of rod BC.

(aBC)n=ω2rCE

Here, (aBC)n is the centripetal acceleration or rod BC.

Draw the free body diagram of the system.

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 16.2, Problem 16.89P , additional homework tip  2

Write the expression for moment about point C.

WABLAB2=IABα+IBCα+rCD(WCDg)(aAB)t+rCE(WBCg)(aBC)t ...... (6)

Write the expression for the equilibrium of forces in horizontal direction.

Cx=(WABg)(aAB)tcosβ+(WABg)(aAB)nsinβ+(WBCg)(aBC)t ...... (7)

Here, Cx is the horizontal reaction at point C.

Write the expression for equilibrium of forces in vertical direction.

CyWABWBc=(WABg)(aAB)tsinβ+(WABg)(aAB)ncosβ+(WBCg)(aBC)n ...... (8)

Here, Cy is the vertical reaction at point C.

Write the expression for resultant reaction.

CR=Cx2+Cy2 ...... (9)

Here CR is the resultant reaction at point C.

Write the expression for the direction of resultant reaction.

ϕ=tan1(CyCx) ...... (10)

Here, ϕ is the direction of resultant reaction at point C.

Calculation:

Substitute 2 lb for WAB, 32.2ft/s2 for g and 1 ft for LAB in equation (1).

IAB=(2)(1)212(32.2)=5.175×103 slugft2

Substitute 4 lb for WBC, 32.2ft/s2 for g and 2 ft for LBC in equation (2).

IBC=(4)(2)212(32.2)=41.4×103 slugft2

Substitute 2 ft for LBC and 0.5 ft for LBD in equation (3).

rCD=(2)2+(0.5)2=2.0615ft

Substitute 2ft for LBC in equation (4).

rCE=22=1 ft

Substitute 0.5 ft for LBD and 2 ft for LBC in equation (5).

tanβ=0.52β=tan1(14)β=14.0362°

Substitute 2 lb for WAB, 1 ft for LAB, 5.175×103 slugft2 for IAB, 41.4×103 slugft2 for IBC, 1ft for rCE, 2.0615ft for rCD, 32.2 ft/s2 for g, αrCD for (aAB)t and αrCE for (aBC)t in equation (6).

2(1)2=[(5.175×103)α+(41.4×103)α+(2.0615)(232.2)(αrCD)+(1)(432.2)(αrCE)]

Substitute 1ft for rCE and 2.0615ft for rCD in above expression and simplify for α.

1=[(5.175×103)α+(41.4×103)α+(2.0615)(232.2)(α(2.0615))+(1)(432.2)(α(1))]1=0.434αα=2.3 rad/s2

Substitute 2 lb for WAB, 4 lb for WBC, 32.2 ft/s2 for g, αrCD for (aAB)t, ω2rCD for (aAB)n, αrCE for (aBC)t and 14.0362° for β in equation (7).

Cx=(232.2)αrCD(cos14.0326°)+(232.2)(rCDω2)sin14.0326°+(432.2)αrCE

Substitute 1ft for rCE, 2.0615ft for rCD, 2.3 rad/s2 for α and 10rad/s for ω in above expression.

Cx=[(232.2)(2.3)(2.0615)(cos14.0326°)+(232.2)((2.0615)(10)2)sin14.0326°+(432.2)(2.3)(1)2]=3.677 lb

Substitute 2 lb for WAB, 4 lb for WBC, 32.2 ft/s2 for g, αrCD for (aAB)t, ω2rCD for (aAB)n, αrCE for (aBC)t

1ft for rCE, 2.0615ft for rCD, 2.3 rad/s2 for α, 10rad/s for ω and 14.0362° for β in equation (8).

Cy24=[(32.2)(2.0615)(2.3)sin(14.0362°)+(232.2)(2.0615)(10)2cos14.0362°+(432.2)(1)(10)2]Cy6=24.773Cy=30.773lb

Substitute 30.773lb for Cy and 3.677lb for Cx in equation (9).

CR=(3.677)2+(30.773)2CR=30.9918lb

Substitute 30.773lb for Cy and 3.677lb for Cx in equation (10).

ϕ=tan1(30.7733.677)ϕ=83.18°

The reaction at point C is 30.9918 lb in the direction of 83.18° measured form positive x -axis.

Conclusion:

Thus, the reaction at point C is 30.9918 lb in the direction of 83.18° measured form positive x -axis.

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Chapter 16 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

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