Question

Dynamics:

Answer 1

.

Answer 2

I think that we can apply the subsequent formula:
\[Pt = mV\]
where \(P\) is the weifìght of the beam, \(V\) is the final speed, \(t\) is the requested time, and \(m\) is the mass of the beam

Answer 3

I did so, got t = 0.317 s , is that correct ?

Answer 4

I got \(t=0.3125\) seconds

Answer 5

Did you do the same steps I did ?

Steps:

P ( resultant of the tension in 2 cables) = sqrt( 2T^2 + 2 T^@Cos2theta)

theta = tan^-1(3/4)

got P = 9899.5

then used the impusle

9899.5t - 5000 t = 155.27 * 10 ( g= 32.2 ft/s^2)

t= 0.317 s ?

Answer 6

no, I have applied a different method, and it is wrong. I think that yours is the correct one

Answer 7

what did you do then ?

Answer 8

my method is too simplified, and doesn't take account of the maximum tension

Answer 9

Thanks michele :D

Answer 10

:)