# Pre Calc

posted by on .

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

• Pre Calc - ,

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 - ,

the vector is actually <2,2>
and the line angle is 64.3

• Pre Calc - ,

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 - ,

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