# Force and potential energy of a mail bag

1. Nov 8, 2007

### physstudent1

1. The problem statement, all variables and given/known data
A 115 kg mail bag hangs by a vertical rope 3.8m long. A postal worker then displaces the bag to a position of 2.2m sideways from its original position, always keeping the rope taut.

What horizontal force is necessary to hold the bag in the new position?
2. Relevant equations

3. The attempt at a solution

Ok, so I set up the new position of the bag and used Pythagoreans theorem to find that the bag is now .7m higher than it was where it started. I called the original position my reference line so it has no potential energy there. At a height of .7 it has .7*9.8*120
In my book it says that F(in x direction) = -(change in potential)/(change in x)

so I thought dividing this potential energy I found by the change in the x direction (2.2) would give me the Force in the horizontal?

Last edited: Nov 8, 2007
2. Nov 9, 2007

### rl.bhat

Draw the diagram as explained in the problem. Draw three vectors. One tension, second horizontal force and third weight vertically downwards. Draw a right angled triange such that length of the rope is hypotenuse, horizontal displacement is horizontal side. According to your calculation the third side becomes 3.1m. Applying the law of trianglesof vectors we can wright F/2.2 = Mg/3.1. Now find F. You can also solve this problem by applying law of moments.