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Question

In the four-bar mechanism shown in figure ABC is the coupler link and all the dimensions are in cms. For the configuration shown, AC is aligned with PA while B is aligned with QB. Then the ratio of instantaneous angular velocity of the link QB to that of the link PA is

A
1.5
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B
0.75
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C
1
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D
1.33
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Solution

The correct option is C 1
ωPA=VAPA

ωQB=VBQB

ΔACB=right angle triangle

ωQBωPA=VBQB×PAVA=VBVA×(PAQB)


The component of velocities of A and B along AB should be same

VAcosθ=VBsinθ

VBVA=1tanθ

tanθ=68=34(from ΔABC tanθ=ACBC)

VBVA=43

ωQBωPA=VBVA×PAQB

=43×2432=1

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