The uniform 888-mm link AB slides up on frictionless rollers as shown below. At the instant shown, the link \( A B \) is vertical and the absolute velocity of point \( A \) is \( 16.7 \mathrm{~m} / \mathrm{s} \) in the direction shown. Ignore the link thickness and roller diametres. For this configuration, \( \theta_{1}=38 \) degrees and \( \theta_{2}=25 \) degrees as shown. Calculate the absolute velocity of the centre of mass at \( G \). Give your answer in the form \( Z=2 v_{G x}+ \) \( \mathrm{v}_{\mathrm{G}_{Y}} \) in \( \mathrm{m} / \mathrm{s} \) where \( \mathrm{v}_{\mathrm{Gx}} \) is the horizontal component of the velocity vector (positive to the right) and \( \mathrm{v}_{\mathrm{Gy}} \) is the vertical component of the velocity vector (positive upwards). Round your final answer to
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First, we need to find the velocity of point B. Since point A is moving at 16.7 m/s at an angle of 38 degrees, we can find the horizontal and vertical components of its velocity: v_Ax = 16.7 * cos(38) = 13.1 m/s (to the right) v_Ay = 16.7 * sin(38) = 10.4 m/s Show more…
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