Okay, I got rid of my post because I'm obviously retarded and not as cool as the dude below me. Listen to that dude.
Find the work done by a force of magnitude of 3 pounds applied in the direction of
v =2i + 2j to move an object 5 feet from to (0, 0) (3, 4)?
We just started learning vectors and nothing seems familiar to me in this problem. Please guide me through step by step how to approach, thanks.
1. Since the direction of the force and the direction of the movement differ by 8.13° the work, which must be done, has to be larger than 15ft-lb.
2. Google for dot-product of vectors: How it is calculated and how the result can be interpreted.
We learned how dot product of vectors is calculated: v(dot product)w=||v||*||w||*cos(theta)
How did you find out that direction of force and the direction of the movement differ by 8.13 degrees? I'm still lost about this problem.
1. You have made a sketch I presume (see attachment)
2. Split the force vector into one vector acting in the dirction of (3, 4) and one which acts perpendicular to this direction (a solid would be lifted off).
3. Use the slopes of the 2 given vectors to determine the included angle.
The work done by a (constant) force F along a vector d is given by (dot product):Originally Posted by metelskiy
W = F d = Fx * dx + Fy * dy (in two dimensions).
Fx, Fy, dx, dy - are components of F and d.
In your problem:
F = 3 pounds (the magnitude)
d = 5 feet (the magnitude)
(since the direction of F is: 2i + 2j)
On the other hand we know that:
From the first equation (substitute dx = (3/4) dy) you get: dy = 4 feet.
Hence: dx = 3 feet. From the second equation you get: Fx = Fy = 2.12.
You can now use these results in the equation for W and you get:
W = 14.8 (pound * feet).
(I am not familiar with these units, we use SI units).