# Thread: Work done by force (vector problem)

1. ## Work done by force (vector problem)

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.

2. ## Re: Work done by force (vector problem)

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.

3. ## Re: Work done by force (vector problem)

Originally Posted by icedtrees
...
It seems to me as if there is excessive information given in the question. <-- no

Work done = Force * Distance <-- This equation is only correct if the direction of the force and the direction of the movement are equal
Distance = 5 feet
Force = 3 pound-force

...
Originally Posted by metelskiy
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.

4. ## Re: Work done by force (vector problem)

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.

5. ## Re: Work done by force (vector problem)

Originally Posted by metelskiy
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.

6. ## Re: Work done by force (vector problem)

Originally Posted by earboth
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.
I'm still not seeing how to achieve the final answer. I know I need length of Force vector and length of direction vector and angle between them. SO is 8.13degrees is the angle between f and d?

7. ## Re: Work done by force (vector problem)

Originally Posted by metelskiy
I'm still not seeing how to achieve the final answer.
The work done by a (constant) force F along a vector d is given by (dot product):
W = F d = Fx * dx + Fy * dy (in two dimensions).
Fx, Fy, dx, dy - are components of F and d.
F = 3 pounds (the magnitude)
d = 5 feet (the magnitude)
$F_x = F_y$ (since the direction of F is: 2i + 2j)
$d_x / d_y = (3 - 0) / (4 - 0) = 3/4 --> d_x = (3/4) d_y.$

On the other hand we know that:
$d_x^2 + d_y^2 = d^2$
$F_x^2 + F_y^2 = F^2$

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).

8. ## Re: Work done by force (vector problem)

Originally Posted by metelskiy
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.
for two vectors expressed as components ...

$\vec{u} = a\vec{i} + b\vec{j}$

and

$\vec{v} = c\vec{i} + d\vec{j}$

the dot product is simply ...

$\vec{u} \cdot \vec{v} = ac + bd$

change the force vector to component notation ...

$\vec{F} = \frac{3}{\sqrt{2}} \vec{i} + \frac{3}{\sqrt{2}} \vec{j}$

the displacement vector was given ...

$\vec{r} = 3\vec{i} + 4\vec{j}$

finally ...

$\vec{F} \cdot \vec{r} = \frac{3}{\sqrt{2}} \cdot 3 + \frac{3}{\sqrt{2}} \cdot 4 = \frac{21}{\sqrt{2}} \, \,$ ft-lbs