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Example: A frictionless pin at A
supports a uniform bar AB of length L as shown in the figure
below. C denotes the center of mass of the
bar. The bar is released from rest
at an angle θ.
Find
the support reactions at A at this instant.
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Strategy: The first
step in applying Euler’s first and/or second law is to draw a free body diagram
to
identify all forces acting on the bar as shown below. Note:
et x en = k
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Apply Euler’s first law F =
m aC
which results in
(Ft + mg sinθ)
et + (Fn – mg cosθ)
en = m
[ act et +
vC2/ρ en ] Since
the bar is released
from rest vC= 0 which gives Ft
+ mg sinθ
= m act and
Fn – mg cosθ =
0
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Next
apply Euler’s 2nd law
in the form MA = IzzA
α + rAC
x m aA Now
|aA|
= 0 so
MA = IzzA α for the bar IzzA =
(1/3) mL2
so mg(L/2)sin θ k = (1/3) mL2
α k
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Thus α
= (3/2)(sin θ)(g/L) . Next use kinematics to express ac in terms of α.
aC = aA –
(v2/ρ) en
+ α k x (– L/2)en Both aA =
0 and
vC = 0 so
aC = (L/2) α et = (L/2)(3/2)(sin θ) (g/L) so Ft
+ mg sinθ
= m [ (3/4) g sin θ ]
Solving
for Ft gives Ft = (¾)
mg sin θ – mg sinθ = – ¼ mg sin θ and
Fn = mg cosθ
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