Expert: Daniel Mazur - 9/18/2005

Question
a man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys.show that the force exerted by each friend on the rope increases as the man moves up.find the force when the man is at a depth of h.  

Answer
I need to know the level of your physics education to give you an answer that fits you. One thing to start with, this is not a "Newton's Laws of Motion" problem, this is a "Newton's Gravitation Law" problem. Secondly, the width d has nothing to do with the problem, it's a redundant information. The pulleys are just dividing the force in half per each friend.

I will respond in general terms as to how to approach the problem.

1) Show that the force increases:

The key is to show, that the NET gravitational force exerted on you BELOW the surface of a planet (R..planet's radius, h..your distance from planet's surface) is equal to the TOTAL gravitational force exerted by a smaller planet of the radius equal to (R-h). This is much easier described using a drewing. Without this, you'll have to try and follow my words.

Draw yourself a ball. Divide the ball of a planet (radius R) into two sections: a smaller ball (radius (R-h)) and a spherical "peel" of thickness h. You need to show that the net grav.force of the "peel" exerted on you, when you are located in the void inside or on its inner surface, is identically zero.

The way to do that is to draw a straght line through your position and create two cones of equal apical spatial angle, symmetrical about the line and having your position at their apex points (they are going to be cones with somewhat crooked bases, but with a bit of differential thinking you'll see it does not have to disturb us). Now, keeping the apical angle fixed (!), your distance to the base of each cone and the areas of the respective bases are interrelated. Should "x" and "y" be your distances from both intercepts of your line and the "peel", the areas of the bases of the above-mentioned cones are
"Sx" proportional to x^2 and "Sy" prop.to y^2.      (*1)

The best way to think about this is choosing an infinitesimal spatial angle, which results in infinitesimal areas. Then you have not difficulties with curvature of the peel and finding the centers of mass of the two bases.

Since the "peel" thickness and material density are assumed constant, the masses of the cone bases have the same proportionality, i.e.
"Mx" prop.to x^2 and "My" prop to y^2.              (*2)

The gravitational forces have proportionality
"Fx" prop.to Mx/(x^2) and "Fy" prop.to My/(y^2).    (*3)

Substitute (*2) to (*3) and observe that Fx and Fy are:
a) independent of x and y
b) of equal magnitude and opposite direction.

And since the points a) and b) are valid independently of our position inside the "peel", choice of apical angle or the direction of the symmetry line of the cones (you can prove this statement, if you want), this leads to the zero net grav.force of the peel.

Considering the attracting part of the mass of the planet is ((R-h)/R)^3, the magnitude of a gravitational force on an object in a creavase goes as (R-h), which is a function rising as h drops.

2) Find the force:

We have done most of the job in the previous part. The result is

F = (1/2) * (-G) * ((R-h)/R)^3 * Mplanet * Mman / (R-h)^2 =
 = (1/2) * (-G) * Mp * Mman * (R-h) / R^3           (*4)

The factor (1/2) comes from the fact that there are two friends sharing the burden. Signs are a matter of convention.

I hope this written form could be understood and helped. Drawing a picture helps a lot.

Good luck!
Daniel