Solutions for Problem Set 6

IX.

Place a force of W_L = Mg acting down on the mass at the left, a force W_R = 2Mg acting down on the right, and since the heavier mass will accelerate downward, a frictional force F_f acting leftward on the middle mass.

X. The force of friction, given mu_s = 0.02 and N = 75 kg * g = 735 Newtons can be as high as 14.7 Newtons. Since this number exceeds the applied force of 1 Newton, the skier does not accelerate.

XI. At any given place on the ramp the forces and geometry are:

Note that tan(theta) = dy/dx since the surface is really curved. I have shown only the tangent line in drawing the local geometry.

Using forces:

y : N cos(theta) - mg = 0

x : N sin(theta) = mv² /x

The velocity is given by v = 2 pi x / T

and theta is given by tan(theta) = dy/dx

Known:	T, x, g
To find:	y(x), or at least tan(theta) = dy/dx
Unknown and unwanted:	N, m
Supposedly irrelevant:	v

We can eliminate the unknown N by dividing the second equation by the first; in the process we happily eliminate m as well:

tan(theta) = v² /(gx)

Eliminating v in favor of T with the last equation [we want to do this since T is a known constant and v is not known or wanted]:

tan(theta) = 4 pi2 x /( gT2 )

It is surprisingly easy to turn this into y(x):

dy/dx = A x

with

A = 4 pi² /( gT² ).

Integrating both sides with respect to x,

y + C = A integral ( x dx ) = A x² / 2

so

y = 2 pi² x² / (gT² ) - C

for any value of C .