Exercício de Dinamica Aplicada (256)

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417 
 
 
PROBLEM 12.72 
The velocity of block A is 2 m/s to the right at the instant when 0.8 mr = 
and 30 .θ = ° Neglecting the mass of the pulley and the effect of friction 
in the pulley and between block A and the horizontal surface, determine, 
at this instant, (a) the tension in the cable, (b) the acceleration of block A, 
(c) the acceleration of block B. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
Let r and θ be polar coordinates of block A as shown, and let By be the 
position coordinate (positive downward, origin at the pulley) for the 
rectilinear motion of block B. 
Radial and transverse components of .Av 
Use either the scalar product of vectors or the triangle construction shown, 
being careful to note the positive directions of the components. 
 
cos30
2cos30 1.73205 m/s
r A r Ar v v= = ⋅ = − °
= − ° = −
v e
 
2
sin 30
2sin30 1.000 m/s
A Ar v vθ θθ = = ⋅ = − °
= ° =
v e
 
1.000
1.25 rad/s
0.8
v
r
θθ = = = 
Constraint of cable: constant,Br y+ = 
 0, 0 or B B Br v r a r a+ = + = = −   (1) 
For block A, : cos or secx A A A A A AF m a T m a T m aθ θΣ = = = (2) 
For block B, + :y B B B B BF m a m g T m aΣ = − = (3) 
Adding Eq. (1) to Eq. (2) to eliminate T, secB A A B Bm g m a m aθ= + (4) 
Radial and transverse components of .Aa 
Use a method similar to that used for the components of velocity. 
 2 cosr A r Ar r a aθ θ− = = ⋅ = −a e (5) 
Using Eq. (1) to eliminate r and changing signs gives 
 2cosB Aa a rθ θ= −  (6)