# The weight of the airliner is 1.20 × 10^6 N and there is an aerodynamic lift force of 1.39 × 10^6 N acting at 30° to the left of the vertical.By drawing a scale vector diagram, show that the...

The weight of the airliner is 1.20 × 10^6 N and there is an aerodynamic lift force of 1.39 × 10^6 N acting at 30° to the left of the vertical.

By drawing a scale vector diagram, show that the resultant of these two

forces.

Please help me cause I am unable to draw this...

Thanks a lot

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Hi, muskan.

I am confident the calculations are correct. I thought I would show you another way to draw the resultant. I included it in the attachment.

One way to consider drawing the vectors is "head to tail". As in, first draw the vector for the weight of the plane, so straight down. Then, from the bottom of the tip of that vector, you draw the vector acting upward on the angle, as I showed in my drawing. The space between the point you started with and the final tip of the arrow of the final vector would be the resultant.

Advantage to this method, pretty easy to remember how to do. Disadvantage, if freehanded and units are poorly approximated, one can't quite tell where the vector stops. Like, in my picture, I have the resultant acting below the horizontal. In the other tutor's picture, they have the resultant acting above the horizontal.

I would prefer the other tutor's way. However, what I showed can provide some insight on some problems, though. Such as, if you aren't given enough information and need some sort of visual, you would at least be able to draw something.

Later, muskan. I hope this helps.

Till Then,

Steve

F1 = 1.39 × 10^6 N acting at 30°

F1x = -1.39 × 10^6 * sin(30) = -0.695 x 10^6 N

F1y = 1.39 × 10^6 * cos(30) = 1.204 x 10^6 N

F2 = 1.20 × 10^6 N

F2x = 0

F2y = -1.20 x 10^6 N

Fx = F1x + F2x = -0.695 x 10^6 + 0 = -0.695 x 10^6 N

Fy = F1y + F2y = 1.204 x 10^6 - 1.20 x 10^6 = 4 x 10^3 N