 #1
shinobi20
 271
 20
This is a problem from Mechanics by Kleppner and Kolenkow1. Homework Statement
Two identical masses M are pivoted at each end of a massless pole of length L. The pole is held leaning against frictionless surfaces at angle θ, as shown, and then released. Find the initial acceleration of each mass.
y^2+x^2=L^2
yy''+(y')^2+xx''+(x')^2=0, if the masses are at initially at rest then yy''+xx''=0, x'' = (y/x)y'' = tan(θ) y''.
The forces that act on the upper mass is the weight (Mg), normal force (Ny) , and force of the pole (F) so the equations are:
My'' = MgFsin(θ) and Ny = Fcos(θ) (No use, can be omitted)
The forces that act on the lower mass is the weight (Mg), normal force (Nx), force of the pole (F) and friction (f) so the equations are:
Mx'' = Fcos(θ)f and NxFsin(θ)Mg=0
By manipulation, I ended up getting y'' = (g)/(1+µtan(θ)+(tan(θ))^2) and so x'' can easily be derived.
[/B]
My question here is that K&K didn't include Fsin(θ) in the second equation of the lower mass. This will result to Nx=Mg which then leads to a different answer. Why did K&K exclude Fsin(θ) knowing that it is a force from the pole that contributes to the downward force for the lower mass?
Two identical masses M are pivoted at each end of a massless pole of length L. The pole is held leaning against frictionless surfaces at angle θ, as shown, and then released. Find the initial acceleration of each mass.
Code:
Note:

o
 \
 \
 \
 \
_____o____
Homework Equations
y^2+x^2=L^2
yy''+(y')^2+xx''+(x')^2=0, if the masses are at initially at rest then yy''+xx''=0, x'' = (y/x)y'' = tan(θ) y''.
The forces that act on the upper mass is the weight (Mg), normal force (Ny) , and force of the pole (F) so the equations are:
My'' = MgFsin(θ) and Ny = Fcos(θ) (No use, can be omitted)
The forces that act on the lower mass is the weight (Mg), normal force (Nx), force of the pole (F) and friction (f) so the equations are:
Mx'' = Fcos(θ)f and NxFsin(θ)Mg=0
The Attempt at a Solution
By manipulation, I ended up getting y'' = (g)/(1+µtan(θ)+(tan(θ))^2) and so x'' can easily be derived.
[/B]
My question here is that K&K didn't include Fsin(θ) in the second equation of the lower mass. This will result to Nx=Mg which then leads to a different answer. Why did K&K exclude Fsin(θ) knowing that it is a force from the pole that contributes to the downward force for the lower mass?
Last edited by a moderator: