Pendulum problem: Horizontal force and work done

[DukeJP2010]

Homework Statement

A simple pendulum (sting has no mass, no air resistance) has a weight on the end of mass m (unknown). The string has a length A horizontal (and always horizontal) force, F, acts on the weight, but it moves so slowly that acceleration is 0.

a)What is the work done by force F?
b)What is the work done by Fg (Fg=mg)? What is the work done by Ft (tension in the string)?

Homework Equations

Work = Fd

The Attempt at a Solution

Here is my attempt at part a. If i am right with a, then i can do 1/2 of b (the mg part) and will attempt Ft after that.

First, break it down into the X and Y components.

Y: FtCos($$\theta$$)-mg=0
X: F-FtSin($$\theta$$)=0

use the Y equation to solve for Ft:
FtCos($$\theta$$)-mg=0
FtCos($$\theta$$)=mg
Ft=mg/(Cos($$\theta$$))

plug Ft into the X equation:
F-FtSin($$\theta$$)=0
F-[mgSin($$\theta$$)]/(Cos($$\theta$$))=0
F=mgtan($$\theta$$)

Now that we got an equation for force, we need to find the distance that the mass moves in the X direction, which is:
dx=lSin($$\theta$$)

so, plug this into W=Fd and get:
W=[mgtan($$\theta$$)][lsin($$\theta$$)]
W=mglTan($$\theta$$)sin($$\theta$$)

Does this look right? If it is right, do pretty much the same thing for Fg. Any clues for work done by Ft?

 

[DukeJP2010]

Ok, since the direction of travel of the mass is always perpendicular to Ft, then Ft can't do any work.

Does the rest of it look OK?

 

[m2003]

Your solution for part a looks correct. For part b, you can use the same approach to find the work done by Fg. Since the acceleration is 0, the tension in the string (Ft) will also be 0. Therefore, the work done by Ft will also be 0.