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Pole leaning against wall - Kleppner

  • #1
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Hello, all. I've just been working on Kleppner to touch up on my knowledge of mechanics and general problem solving skills and I'm running into a bit of trouble on this leaning pole one in chapter 2.

1. Homework Statement

A pole of negligible mass leans against a wall, at angle θ with the horizontal. Gravity is directed down.

(a) Find the constraint relating the vertical acceleration of one end to the horizontal acceleration of the other.

(b) Now suppose that each end carries a pivoted mass M. Find the initial vertical and horizontal components of acceleration as the pole just begins to slide on the frictionless wall and floor. Assume that at the beginning of the motion the forces exerted by the rod are along the line of the rod. (As the motion progresses, the system rotates and the rod exerts sidewise forces.)


The Attempt at a Solution



For part a) I used the fact that the length of the pole is going to be constant and so

[tex]

x^2+y^2=L^2 \\ 2xx'+2yy'=0 \\ xx'+yy'=0 \\

[/tex]

Taking the second time derivative,

[tex]

\left(x'\right)^2+xx''+\left(y'\right)^2+yy''=0 \\

[/tex]

Because the pole is at rest, we can let the velocity terms be zero and so we have,

[tex]

xx''=-yy'' \\

x''=-tan\left(\theta \right)y''

[/tex]

Now for part b). After we attach the masses to the ends.

The forces acting on the mass that is along the y-axis, is the gravitational force and the normal force because it's against the wall. The normal force pushing it against the wall must be due to the pole and so it is acting at an angle theta to the mass. The vertical component of this force is given by sine of theta. That is,

[tex]
my''\:=\:F_psin\left(\theta \right)-mg
[/tex]

The forces acting the mass on the ground are the gravitational force, the normal force, and the force by the pole. In this case because the mass is resting flat on the ground, the normal force will cancel out it's weight, and so the net force acting on it is just the force due to the pole. That is,

[tex]
mx''\:=\:F_psin\left(\theta \right)
[/tex]

Solving for y'' and x'' using the first constraint equation

[tex]
my''\:=\:mx''-mg\\y''=x''-g \\ y''\:=-tan\left(\theta \right)y''-g\\ \\~\\ y''=-\frac{g}{1+tan\left(\theta \right)}
\hspace{1cm}
x''=\frac{gtan\left(\theta \right)}{1+tan\left(\theta \right)}
[/tex]

However, the answers given are (below), and I for the life of me can't figure out can't see what I'm doing wrong. I even reluctantly studied the full solution and for whatever reason still can't make sense of it / what I'm doing wrong. Any help is much appreciated.

[tex]

y''\:=\:-gcos^2\left(\theta \right)
\hspace{1cm}
x''\:=\:gcos\left(\theta \right)sin\left(\theta \right)

[/tex]
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
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For part a, there is an easier way. Just consider the components of acceleration along the rod.
mx′′=Fpsin(θ)
You might care to reconsider that.
 

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