VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter 16.2, Problem 16.134P

The hatchback of a car is positioned as shown to help determine the appropriate size for a damping mechanism AB. The weight of the door is 40 lbs, and its mass moment of inertia about the center of gravity G is 15 lb·ft·s2. The linkage DEFH controls the motion of the hatch and is shown in more detail in part (b) of the figure. Assume that the mass of the links DE, EF, and FH are negligible, compared to the mass of the door. With AB removed, determine (a) the initial angular acceleration of the 40-lb door as it is released from rest, (b) the force on link FH.

Chapter 16.2, Problem 16.134P, The hatchback of a car is positioned as shown to help determine the appropriate size for a damping , example  1

(a)

Chapter 16.2, Problem 16.134P, The hatchback of a car is positioned as shown to help determine the appropriate size for a damping , example  2

(b)

Fig. P16.134

(a)

Expert Solution
Check Mark
To determine

The initial angular acceleration of the door.

Answer to Problem 16.134P

The initial angular acceleration of the door is αEF=4.144rad/s2_.

Explanation of Solution

Given information:

The weight of the door is W=40lbs.

The mass moment of inertia of the center of gravity is IG=15lbfts2.

Calculation:

Consider the acceleration due to gravity as g=32.2ft/s2.

Calculate the mass (m) as shown below.

m=Wg

Substitute 40lbs for W and 32.2ft/s2 for g.

m=4032.2=1.242lbs2/ft×1slugs1lbs2/ft=1.242slugs

Sketch the Free Body Diagram of the door as shown in Figure 1.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 16.2, Problem 16.134P

Refer to Figure 1.

Calculate the distance (a) as shown below.

a=2cos24°=1.827in.×1ft12in.=0.1523ft

Calculate the distance (b) as shown below.

b=2sin24°=0.8135in.×1ft12in.=0.06779ft

Calculate the position vectors as shown below.

The position of F with respect to H.

rF/H=(4in.×1ft12in.)j=13j

The position of E with respect to F.

rE/F=ai+bj

Substitute 0.1523ft for a and 0.06779ft for b.

rE/F=0.1523i+0.06779j

The position of E with respect to D.

rE/D=(9.5in.×1ft12in.)cos48°i+(9.5in.×1ft12in.)sin48°j=0.5297i+0.5883j

The position of G with respect to E.

rG/E=2i

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium of forces along x direction as shown below.

Fx=maxFEcos48°=ma¯x

Substitute 1.242slugs for m.

0.6691FE=1.242a¯xFE=1.856a¯x (1)

Apply the Equilibrium of forces along y direction as shown below.

Fy=maymg+FEsin48°+FF=ma¯y

Substitute 1.242slugs for m and 32.2ft/s2 for g.

1.242×32.2+FEsin48°+FF=1.242a¯y40+0.743FE+FF=1.242a¯y

Substitute 1.856a¯x for FE.

40+0.743×1.856a¯x+FF=1.242a¯yFF=1.242a¯y1.379a¯x+40 (2)

Apply the Equilibrium of moment about G as shown below.

MG=I¯αEF2.1523FF+0.06779FEcos48°+2FEsin48°=15αEF2.1523FF+1.53165FE=15αEF (3)

Calculate the acceleration at F (aF) as shown below.

aF=aHωFH2rF/H+αFH×rF/H

Substitute 0 for aH, 0 for ωFH, and 13j for rF/H.

aF=0+αFHk×13j=13αFHi

Calculate the acceleration at E (aE) as shown below.

aE=aFωEF2rE/F+αEF×rE/F

Substitute 13αFHi for aF, 0 for ωEF, and 0.1523i+0.06779j for rE/F.

aE=13αFHi+0+αEFk×(0.1523i+0.06779j)=13αFHi0.1523αEFj0.06779αEFi=(13αFH+0.06779αEF)i0.1523αEFj (4)

Calculate the acceleration at E (aE) as shown below.

aE=aDωDE2rE/D+αDE×rE/D

Substitute 0 for aD, 0 for ωDE, and 0.5297i+0.5883j for rE/D.

aE=0+0+αDEk×(0.5297i+0.5883j)=0.5883αDEi+0.5297αDEj (5)

Equating Equations (4) and (5) as shown below.

(13αFH+0.06779αEF)i0.1523αEFj=0.5883αDEi+0.5297αDEj

Resolving i and j components as shown below.

For i component.

13αFH+0.06779αEF=0.5883αDE (6)

For j component.

0.1523αEF=0.5297αDEαDE=0.2875αEF

Calculate the relative acceleration (a¯G) as shown below.

a¯G=aEωEF2rG/E+αEF×rG/E

Substitute 0.5883αDEi+0.5297αDEj for aE, 0 for ωEF, and 2i for rG/E.

a¯G=0.5883αDEi+0.5297αDEj+0+αEFk×(2i)a¯xi+a¯yj=0.5883αDEi+0.5297αDEj2αEFja¯xi+a¯yj=0.5883αDEi+(0.5297αDE2αEF)j

Resolving i and j components as shown below.

For i component.

a¯x=0.5883αDE

For j component.

a¯y=0.5297αDE2αEF

Calculate the force at F (FF) as shown below.

Substitute 0.5883αDE for a¯x and 0.5297αDE2αEF for a¯y in Equation (2).

FF=1.242(0.5297αDE2αEF)1.379(0.5883αDE)+40=0.6579αDE2.484αEF+0.8113αDE+40=1.4692αDE2.484αEF+40

Substitute 0.2875αEF for αDE.

FF=1.4692×(0.2875αEF)2.484αEF+40=2.9064αEF+40 (7)

Calculate the force at E (FE) as shown below.

Substitute 0.5883αDE for a¯x in Equation (1).

FE=1.856×(0.5883αDE)=1.0919αDE

Substitute 0.2875αEF for αDE.

FE=1.0919×(0.2875αEF)=0.3139αEF

Calculate the angular acceleration (αEF) as shown below.

Substitute 2.9064αEF+40 for FF and 0.3139αEF for FE in Equation (3).

2.1523(2.9064αEF+40)+1.53165(0.3139αEF)=15αEF6.2554αEF+86.092+0.4808αEF15αEF=020.7746αEF=86.092αEF=4.144rad/s2

Therefore, the initial angular acceleration of the door is αEF=4.144rad/s2_.

(b)

Expert Solution
Check Mark
To determine

The force on link FH.

Answer to Problem 16.134P

The force on link FH is FF=27.96N_.

Explanation of Solution

Given information:

The weight of the door is W=40lbs.

The mass moment of inertia of the center of gravity is IG=15lbfts2.

Calculation:

Refer to part (a).

The initial angular acceleration of the door is αEF=4.144rad/s2.

Calculate the force at F (FF) as shown below.

Substitute 4.144rad/s2 for αEF in Equation (7).

FF=2.9064×4.144+40=27.96N

Therefore, the force on link FH is FF=27.96N_.

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Chapter 16 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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