Question

Your answer is partially correct. Try again. The ends of the 0.30-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.40 m/s and an acceleration of 0.46 m/s in the directions shown. Determine the angular acceleration (positive if counterdockwise, negative if dockwise) of the bar and the acceleration of end A (positive if up, negative if down). 0.30 m 1101 -0.46 m 30-0.40

Answer 1

First a neat sketch with 40 deg as third angle of triangle

0.30 1010

Take advantage of the constraints (B stays on the floor, A stays on the wall)

Velocity eqn

vA = vB + ω × rA/B

VA = UB + wrA/B

Vai +vAj= (0.4)i + wk(0.3)(-cos40i + sin40j)

0 . 2303- ;( 0 . 192 - 0 . 4) = زد

Since vAi= 0, Equating i components we find

0 = (0.4 -0.1920)

w=2.0833rad/s

Thus angular velocity is obtained

Now for angular acceleration

aA = aB + α × rD/B − ω2rD/B

aA = a +a*ra/B-W-LAB

aaj = api + ak* L(-costi + sinj) - w2L(- cos i + sin 0j)

A] = (ab - aL sin 0 +w²L cos)i + (-aL cos 0 - wºL sin);

Equating components we have

ag = alsing - w2Lcose

-0.46 = a(0.3)sin 40 – (2.0833)2(0.3)cos40

a = 2.786rad/s2

Positive for counterclockwise

Now acceleration of end A

A = a Lcosi - W2 Lsino

aA = 2.786(0.3)cos40 - (2.0833)2(0.3)sin 40

A = -0.1966m/s?

Sign is negative, so downwards.

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