Posted by **Em** on Wednesday, May 25, 2011 at 7:06pm.

Find the work done by a force F of 30 pounds acting in the direction (2,3) in moving an object 3 feet from (0,0) to the point in the first quadrant along the line y = (1/2)x.

The answer is 85.38 foot-pounds.

I just need to figure out how to get the answer.

- Pre Calc -
**Damon**, Wednesday, May 25, 2011 at 7:32pm
Find the angle of the Force F

cos T = 2/sqrt13 = .555

T = 56.3 deg above x axis

or

tan T = 2/3

so T = 56.3

Find the direction of motion from slope of line

tan slope = 1/2

slope angle = 26.6 deg

angle between force and motion = 56.3-26.6 = 29.7 degrees

so

component of force in direction of motion = 30 cos 29.7

= 26.06 pounds

26.06*3 = 78.2 ft lbs

beats me how to get 85.38

It would be much closer if the force vector were direction(3,2)

- Pre Calc -
**Megha**, Wednesday, January 30, 2013 at 9:11pm
the vector is actually <2,2>

and the line angle is 64.3

- Pre Calc -
**Oscar**, Thursday, December 4, 2014 at 10:11pm
the vector is indeed <2,2>, but the line angle (angle formed by y=1/2x and x-axis) is as Damon posted, 26.6 degree. As for why it isn't 64.3 degree is because slope is change in y over the change in x, thus this means horizontal change is 2 and vertical change is 1. When finding the line angle, simply use arc tan(opp/adj) or in this case arc tan(1/2) which will result in 26.6 degree.

As for the answer, its actually 95.36 ft lb if the vector is <2,2>

- Pre Calc -
**Katelyn**, Wednesday, April 29, 2015 at 7:51pm
F1 -> 30 lbs, direction <2, 2>

magnitude of <2, 2> = 2sqrt(2)

Find unit vector of F1 (divide original components by magnitude)

-> <30/sqrt(2), 30/sqrt(2)>

F2 -> 3 feet, direction y = 1/2x

3 = sqrt(x^2 + y^2)

y = 1/2x

3 = sqrt(x^2 + (1/2x)^2)

9 = 5/4* x^2

x^2 = 36/5

x = 6/sqrt(5)

y = 2/sqrt(5)

F1 = <30/sqrt(2), 30/sqrt(2)>

F2 = <6/sqrt(5), 3/sqrt(5)>

Find dot product of F1 and F2

(30/sqrt(2))*(6/sqrt(5))+(30/sqrt(2))*(3/sqrt(5)) = 85.38

Answer in the back of the book is 85.38

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