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
[tex]mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2[/tex]
Limit is t1 to t2
[tex]-(\frac{14}{32.2}) (2) (3) - T = \frac{14}{32.2} v_y_2[/tex] -(1)
Applying the principle of Impulse and momentum in y-direction
[tex]mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2[/tex]
Limit is t1 to t2
[tex](\frac{5}{32.2}) (3) - 2T + 3 = -\frac{5}{32.2} \frac{v_y_2}{2}[/tex] -(2)
Solving equation (1) and (2), we obtain
T = 1.6lb
va2 = 10.5 ft/s
