Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259977251
Author: BEER
Publisher: MCG
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Chapter 16.2, Problem 16.127P

The test rig shown was developed to perform fatigue testing on fitness trampolines. A motor drives the 200-mm radius flywheel AB, which is pinned at its center point A, in a counterclockwise direction with a constant angular velocity of 120 rpm. The flywheel is attached to slider CD by the 400-mm connecting rod BC. The mass of the connecting rod BC is 5 kg, and the mass of the link CD and foot is 2 kg. At the instant when θ = 0° and the foot is just above the trampoline, determine the force exerted by pin C on rod BC.

Chapter 16.2, Problem 16.127P, The test rig shown was developed to perform fatigue testing on fitness trampolines. A motor drives

Fig. P16.127

Expert Solution & Answer
Check Mark
To determine

Find the force exerted by pin C on rod BC for fitness trampoline.

Answer to Problem 16.127P

The force exerted by pin C on rod BC is 67.62N_.

Explanation of Solution

Given information:

The radius of the flywheel AB is r=200mm.

The mass of the connecting rod BC is mBC=5kg.

The mass of the link CD and foot is mCD=2kg.

The length of the rod BC is lBC=400mm.

The angular velocity is ωAB=120rpm.

The angle is θ=0°.

Calculation:

Convert the unit of angular velocity from rpm to rad/s as shown below.

ωAB=120rpm×2π60rad/s1rpm=4πrad/s

Consider the acceleration due to gravity as g=9.81ft/s2.

Calculate the weight (W) as shown below.

W=mg (1)

Calculate the weight of rod BC (WBC) as shown below.

Substitute 5kg for m and 9.81ft/s2 for g in Equation (1).

WBC=5×9.81=49.05kgm/s2×1N1kgm/s2=49.05N

Calculate the weight of link CD (WCD) as shown below.

Substitute 2kg for m and 9.81ft/s2 for g in Equation (1).

WCD=2×9.81=19.62kgm/s2×1N1kgm/s2=19.62N

Sketch the geometry of the rig as shown in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.127P , additional homework tip  1

Refer to Figure 1.

Calculate the angle (α) as shown below.

cosα=0.20.4α=60°

Calculate the position vectors (r) as shown below.

Position of B with respect to A.

rB/A=0.2i

Position of C with respect to B.

rC/B=0.4cos60°i+0.4sin60°j=0.2i+0.3464j

Position of G with respect to B.

rG/B=0.1i+0.13j=0.1i+0.1732j

Calculate the moment of inertia (I¯BC) as shown below.

I¯BC=112mBClBC2

Substitute 400mm for l and 5kg for m.

I¯BC=112×5kg×(400mm×1m1,000mm)2=0.0667kgm2

Sketch the Free Body Diagram of the rod CD as shown in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.127P , additional homework tip  2

Refer to Figure 2.

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

Fy=ma2gCy=mCDa¯CD,y

Substitute 9.81m/s2 for g and 2kg for mCD.

2×9.81Cy=2a¯CD,yCy+2a¯CD,y+19.62=0 (2)

Sketch the Free Body Diagram of rod BC as shown in Figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.127P , additional homework tip  3

Refer to Figure 3.

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

Fx=maCx+Bx=mBCa¯BC,x

Substitute 5kg for mBC.

Cx+Bx=5a¯BC,xBx=5a¯BC,xCx (3)

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

Fy=maCy+BymBCg=mBCa¯BC,y

Substitute 9.81m/s2 for g and 2kg for mCD.

Cy+By5×9.81=5a¯BC,yCy+By5a¯BC,y49.05=0 (4)

Apply the Equilibrium of moment about G as shown below.

MG=IGα0.13Cx0.1Cy+0.13Bx+0.1By=IBCαBC0.1732Cx0.1Cy+0.1732Bx+0.1By=IBCαBC

Substitute 0.0667kgm2 for I¯BC.

0.1732Cx0.1Cy+0.1732Bx+0.1By=0.0667αBC (5)

Calculate the velocity (vB) as shown below.

vB=vA+ωAB×rB/A

Substitute 0 for vA, 4πrad/s for ωAB, and 0.2i for rB/A.

vB=0+4πk×0.2i=0.8πj

Calculate the velocity (vC) as shown below.

vC=vB+ωBC×rC/B

Substitute 0.8πj for vB and 0.2i+0.3464j for rC/B.

vCj=0.8πj+ωBCk×(0.2i+0.3464j)=0.8πj0.2ωBCj0.3464ωBCi=(0.8π0.2ωBC)j0.3464ωBCi

Resolving the i and j components as shown below.

For i component.

0.3464ωBC=0ωBC=0

For j component.

vC=0.8π0.2ωBC

Substitute 0 for ωBC.

vC=0.8πm/s

Calculate the relative acceleration (aB) as shown below.

aB=aAωAB2rB/A+αAB×rB/A

Substitute 0 for aA, 4πrad/s for ωAB, 0.2i for rB/A, and 0 for αAB.

aB=(4π)2×0.2i=3.2π2i

Calculate the relative acceleration (aC) as shown below.

aC=aBωBC2rC/B+αBC×rC/B

Substitute 3.2π2i for aB, 0 for ωBC, and 0.2i+0.3464j for rC/B.

aCj=3.2π2i0+αBCk×(0.2i+0.3464j)=3.2π2i0.2αBCj0.3464αBCi=(3.2π2+0.3464αBC)i0.2αBCj

Resolving i and j components as shown below.

For i component,

3.2π2+0.3464αBC=00.3464αBC=31.5827αBC=91.17rad/s2

For j component,

aC=0.2αBC

Substitute 91.17rad/s2 for αBC.

aC=a¯CD,y=0.2×91.17=18.234m/s2

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

a¯BC=aBωBC2rG/B+αBC×rG/B

Substitute 3.2π2i for aB, 0 for ωBC, 91.17rad/s2 for αBC, and 0.1i+0.1732j for rG/B.

aBC,xi+aBC,yj=3.2π2i0+(91.17)k×(0.1i+0.1732j)=3.2π2i+9.117j+15.7906i=15.79i+9.117j

Resolving i and j components as shown below.

aBC,x=15.79m/s2aBC,y=9.117m/s2

Calculate the reaction (Cy) as shown below.

Substitute 18.234m/s2 for a¯CD,y in Equation (2).

Cy+2×18.234+19.62=0Cy+56.088=0Cy=56.088N

Calculate the reaction (By) as shown below.

Substitute 56.088N for Cy and 9.117m/s2 for a¯BC,y in Equation (4).

56.088+By5×9.11749.05=0By150.723=0By=150.723N

Calculate the reaction (Cx) as shown below.

Substitute 5a¯BC,xCx for Bx, 150.723N for By, 56.088N for Cy, and 91.17rad/s2 for αBC in Equation (5).

[0.1732Cx0.1×(56.088)+0.1732(5a¯BC,xCx)+0.1×150.723]=0.0667×(91.17)0.1732Cx+20.6811+0.866a¯BC,x0.1732Cx=6.0810.3464Cx+0.866a¯BC,x+26.7621=0

Substitute 15.79m/s2 for a¯BC,x.

0.3464Cx+0.866×(15.79)+26.7621=00.3464Cx=13.08796Cx=37.78N

Calculate the reaction at C as shown below.

C=Cx2+Cy2

Substitute 37.78N for Cx and 56.088N for Cy.

C=37.782+(56.088)2=4,573.192144=67.62N

Therefore, the force exerted by pin C on rod BC is 67.62N_.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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