Question

Block A weighs 14 lb and block B weighs 5 lb If B is moving downward with a velocity (vB)1 = 3 ft/s at t = 0, determine the velocity of A when t = 1 s. Assume that the horizontal plane is smooth. Neglect the mass of the pulleys and cords.

Answer 1

Velocity, va2 = 10.5 ft/s

Explanation:

From the figure:

Length of the cable = Sa + 2Sb = l

∴ vₐ = -2vb

Applying the principle of Impulse and momentum in x-direction

$mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2$

Limit is t1 to t2

$-(\frac{14}{32.2}) (2) (3) - T = \frac{14}{32.2} v_y_2$                           -(1)

Applying the principle of Impulse and momentum in y-direction

$mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2$

Limit is t1 to t2

$(\frac{5}{32.2}) (3) - 2T + 3 = -\frac{5}{32.2} \frac{v_y_2}{2}$                            -(2)

Solving equation (1) and (2), we obtain

T = 1.6lb

va2 = 10.5 ft/s