How Accurate Are the Calculations for Plane Speed and Cable Tension?

[WMDhamnekar]

1)An airplane heads due east at 300 mph through a tailwind whose velocity is given by $w=\langle20,20\rangle$

How fast is the tailwind blowing? In what direction? How fast is the plane flying? In what direction?Answer: The tailwind is blowing at 28.2843 mph approx.in $45^{\circ}$ direction north of due east.Plane is flying at 424.2641 mph approx. in $45^{\circ}$ north of due east. Are these answers correct?

2)A 50kg block is suspended by 2 identical cables of the same length forming $45^{\circ}$ angles with the horizontal.

If the block remains motionless over time, then what is the magnitude of the tensile force exerted by each of the cables on the block?

Answer provided is 346.48 Newtons. My answer is 245N. which is correct?

 

[HOI]

Your answer for the tailwind (I presume these are your answers) is correct. The velocity vector for the airplane is <300, 0>+ <20, 20>= <320, 20>. The speed is [tex]\sqrt{320^2+ 20^2}= \sqrt{102800}= 320.62[/tex] mph, approximately. (A 28 mph tailwind will certainly NOT increase an airplanes speed by over 100 mph!) The direction is also NOT the same as the tail wind. The angle (as degrees from true north) is given by 90- arctan(20/320)= 86.42 degrees, less than 4 degrees off true east.

For the second problem, since we have two cables, each at the same angle, each cable will be supporting half of the weight, 25 kg * 9.81= 245.25 Newtons. That is, of course, vertical. Since the cables are at 45 degrees, the horizontal component is the same. The tension in the cable itself is [tex]\sqrt{245.25^2+ 245.25^2}= 245.25\sqrt{2}= 346.84 [/tex] Newtons.