# Microlight Prob

• Dec 1st 2009, 02:10 AM
Microlight Prob
Ok, so im having a slight problem with getting to the same resultant as suggested by my book. Perhaps you could help me...

The microlights cruising speed in still air is 45 ms^-1, pointing in the direction of N 54◦ W but flies in a wind speed of 15.2 ms^-1 from a S 28W. We Take i to be 1 ms^-1 due east and j to be 1 ms^-1 due north.
Vm - Velocity Microlight
Vw - Velocity Wind

So Vm has a magnitude of 45 and direction 28◦
The wind comes from S 28◦ W & hence blows towards N 28◦ E for which the direction is -(90◦ -28◦) hence vector Vm has a magnitude 15.2 and direction -62

The components are therefore

Vm = 45 cos(28◦)i + 45sin(28◦)
= 39.7326i + 21.1262J

Vw = 15.2 cos(-62◦)i + 15.2cos (-62)J
=-7.1360i + (-13.4208)J

so i make the resultant

v = Vm + Vw =

39.7326 + (-7.1360)i + 21.1262 + (-13.4208)J

=32.5966i & 7.7054

The suggested resultant is -29.2698i + 39.8711J, so where am i going wrong, thanks guys!
• Dec 1st 2009, 04:30 AM
Soroban

Quote:

The microlight's cruising speed in still air is 45 m/s in the direction of N54°W.
but flies in a wind speed of 15.2 m/s from S28°W.

Find its resultant vector.

The plane's vector has a magnitude of 45 and an angle of 144°.

. . $v_p \:=\;\langle45\cos144^o,\;45\sin144^o\rangle$

The wind's vector has a magnitude of 15.2 and an angle of 62°.

. . $v_w \;=\;\langle 15.2\cos62^o,\;15.2\sin62^o\rangle$

The resultant vector is:

. . $v_p + v_w \;=\;\langle45\cos144^o,\;45\sin144^o\rangle + \langle15.2\cos62^o,\;15.2\sin62^o\rangle$

. . . . . . . $=\;\langle -29.26979699,\;39.87113976 \rangle$

• Dec 1st 2009, 04:33 AM
Quote:

Ok, so im having a slight problem with getting to the same resultant as suggested by my book. Perhaps you could help me...

The microlights cruising speed in still air is 45 ms^-1, pointing in the direction of N 54◦ W but flies in a wind speed of 15.2 ms^-1 from a S 28W. We Take i to be 1 ms^-1 due east and j to be 1 ms^-1 due north.
Vm - Velocity Microlight
Vw - Velocity Wind

So Vm has a magnitude of 45 and direction 28◦
The wind comes from S 28◦ W & hence blows towards N 28◦ E for which the direction is -(90◦ -28◦) hence vector Vm has a magnitude 15.2 and direction -62

The components are therefore

Vm = 45 cos(28◦)i + 45sin(28◦)
= 39.7326i + 21.1262J

Vw = 15.2 cos(-62◦)i + 15.2cos (-62)J
=-7.1360i + (-13.4208)J

so i make the resultant

v = Vm + Vw =

39.7326 + (-7.1360)i + 21.1262 + (-13.4208)J

=32.5966i & 7.7054

The suggested resultant is -29.2698i + 39.8711J, so where am i going wrong, thanks guys!

Thanks for showing us all your working, but I'm afraid I don't understand how you arrived at the angles you used in your calculations. $v_m$ is in a direction N $54^o$ W, so its components are:
$-45\sin 54^o$ East and $45\cos54^o$ North
So:
$v_m = -36.41\textbf{i}+26.45\textbf{j}$
And the wind (as you said) blows in a direction N $28^o$ E, so its components are:
$15.2\sin28^o$ East and $15.2\cos28^o$ North
So:
$v_w = 7.14\textbf{i}+13.42\textbf{j}$
which gives a resultant velocity of:
$-29.27\textbf{i}+39.87\textbf{j}$
agreeing with the answer in the book.

• Dec 1st 2009, 05:16 AM
Quote:

Originally Posted by Soroban

The plane's vector has a magnitude of 45 and an angle of 144°.

. . $v_p \:=\;\langle45\cos144^o,\;45\sin144^o\rangle$

Thank you - so i was nearly there, all apart from the angle, so how is this derived?

Thanks
• Dec 1st 2009, 05:43 AM
Soroban

The angle of a vector is always measured from the positive $x$-axis.
(Counter-clockwise: positive angle. .Clockwise: negative angle.)

The plane's direction is N54°W

Code:

                |         *      |           *    |             *54°|               * |       - - - - - * - - - - -                 |

Measured from the positive x-axis, the angle is 144°.
Code:

        *           *             *  144°               *       - - - - - * - - - - -                 |

The wind's direction is from S28°W.
Code:

                 |                 |       - - - - - + - - - - -               /|               / |             /  |             /28°|           /    |                 |

Measured from the positive x-axis, the angle is 62°.

Code:

                |    /                 |28°/                 |  /                 | /                 |/ 62°       - - - - - + - - - - -               /|               / |             /  |             /28°|           /    |                 |