Who can answer this engineering miechanics problem on the resultant of parallel forces??

the 16 ft wing of an airplane is subjected to a lift which varies from zero at the tip to 360 lb per ft at the fuselage according to w=90x^1/2 lb per ft where x is measured from the tip. Compute the resultant and its location from the wing tip.

please answer and explanation..

Answer:
Given the force per unit length w, a "chunk" of this force is

dF=wdx

To find the total force, integrate over the length of the wing, from zero to 16ft

F=int(90x^(1/2)dx)= 90(2/3)x^(3/2) x=0..16, = 3840lbs

The chunk of torque on the wing is given as dT=xdF

Again summing via the integral

T=int(x90x^(1/2)dx)= 90(2/5)x^(5/2), x=0..16, = 36864ft(lbs)

Thus the location of the force found in the first part should give the resultant torque, or 36864ft(lbs)/36864lbs = 9.6ft, so this force would be 9.6ft from the wing tip.

Thanks to Geek below for revealing an error in my arithmetic.

Integrate 90x^1/2 from 0 to 16 to find the resultant force on the wing.

F=90*2/3*(16)^3/2-0
F=3840 lbs

Find the value of x where W=1/2F to find the location of the resultant.

1920=90*2/3(x)^3/2
x^3/2=32
x=10.08 ft from the wing tip

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Who can answer this engineering miechanics problem on the resultant of parallel forces??

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