Normal forces, at an angle

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normal forces, at an angle

i was wondering, if we had a ladder propped up against frictionless wall and makes an angle theta with the ground, and the ground is frictionless too, what would be the magnitude of the normal forces from the ground and the floor, and how would you calculate the motion of the center of mass of the ladder? it makes a quarter-circle, but how long does it take? (does the center really accelerate down at 9.8m/s/s in this case?)

and, i think the normal forces might just be the sine of the angles that the ladder forms, but if this is so, why?
 

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  • #2
futb0l
Well, you would have to calculate the torques and the forces. Draw a free body diagram.
 
  • #3
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yes, it is a quarter-circle....
don't calculate the torques and force because the whole system is accelerating and the acceleration is unknown... solving the acceleration (linear and circular) has no different with solving your problem... you can use consevation of energy in this case... write down everything in term of theta... if you do it correctly, you will get something like this
[tex] \frac{d\theta}{dt} = K \sqrt{sin\theta} [/tex]

in order to find how the angle depend on time, you need to solve the above DE, which is not solvable...(you can't use sinx=x here since you did not say theta is small)...
if you are in highschool and have no idea how do deal with DE or calculus, let me know...
 
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i dont even see how you got that expression... its something like m*g*l*sin(theta)=horizontal kinetic energy and rotational kinetic energy.

how do you derive that? yes i'm in high school, but i know some calculus.

anyway, is there no way of finding the initial normal forces?
 
  • #5
dextercioby
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vincentchan said:
[tex] \frac{d\theta}{dt} = K \sqrt{sin\theta} [/tex]

in order to find how the angle depend on time, you need to solve the above DE, which is not solvable...(you can't use sinx=x here since you did not say theta is small)...

THAT EQUATION CAN BE SOLVED.It's called SEPARATION OF VARIABLES.The integrator at Wolfram expressed in terms of Legendre Elliptic Integrals of the First Kind.And yes,theta of 't' is given in terms of Jacobi Sinus Amplitudinis...

Daniel.

P.S.Don't mislead people with your lack of knowledge... :rolleyes:

P.P.S. I like the answer when u type the function. :rofl: http://www.calc101.com/webMathematica/integrals.jsp#topdoit
 
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dex:
...... sorry of my lack of knowledge :frown: , Could you show me how to solve this integral... I really wanna know....

daveed:
...... is this a HW problem (I really don't think your teacher will ask you do solve the exact solution) ? or you just curious? The rule here is, we don't do HW for people.....

daveed said:
m*g*l*sin(theta)=horizontal kinetic energy and rotational kinetic energy.

you are basically right, but the translational kinetic energy has two part, the ladder has vertical KE as well as horizontal KE... as I told you before, see how the KE (rotational and translational) depend on the first derivative of [itex] \theta [/itex].... use some basic algebra, and you will get the differential equation I gave you earlier

Edit:
after you know how [itex] \theta [/itex] depend on time, you will get the vertical acceleration, then the normal is just F_total=ma.....
We need to waiting for Dextercioby give us the exact solution for theta, I reallly cant helpp you on that part :cry:
 
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  • #7
dextercioby
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Sorry,i didn't mean to offend...It's this integral:
[tex] \int \frac{dx}{\sqrt{\sin x}} [/tex]
Try any "conventional" methods...It won't work...U won't find the primitive through elementary functions.As i said,it's a typical Elliptic Integral of the First Kind.Read more here
U can use the integrator on their page to find the answer to this specific integral.I believe you can find it in Gradsteyn & Rythzik...

Daniel.
 
  • #8
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its not a homework problem, just a tangent to the equilibrium problems that have friction. but thanks for the help. my "basic algebra" is not good enough to get the differential equation you gave me, however. =(
 
  • #9
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the site talked about [tex] \int \frac{dx}{\sqrt{1-k^2 sin x} } [/tex] for 0 < k < 1
but that is not the integral i wanted... ... did i miss something?
 
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  • #10
Gokul43201
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I'm not sure how all you folks get [tex]\dot {\theta} = K \sqrt {sin \theta} [/tex]

I get [tex]\ddot {\theta} = -(3g/2l)cos \theta [/tex]
 
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  • #11
dextercioby
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vincentchan said:
the site talked about [tex] \int \frac{dx}{\sqrt{1-k^{2}\sin^{2}x}} [/tex] for 0 < k < 1
but that is not the integral i wanted... ... did i miss something?

There can be made a substitution which would bring the initial integral to the standard form presented on the site.

Anyway,u have two options:either solve it,or drop it...Besides,you have to invert the EIntegral,and would more p.i.a.

Daniel.

P.S.I think it's more complicated that the mathematical pendulum... :rolleyes:
 
  • #12
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how do you get the second derivative..... I used conservation of energy.... the rotational KE and translational KE is propotional to [itex] \frac{d \theta}{dt} [/itex] square , PE = [itex] mglsin \theta /2 [/itex], set both side equal.... what approach did you use? and one more question, do you think the DE I mention earlier is solvable?
 
  • #13
dextercioby
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He used Newton's law for rotational movement.U used conservation of energy.U sould have gotten somthing else,though...One of you...They don't seem to match.

Yes,both equations have solutions expressible through the Jacobi Elliptic Functions...

Daniel.
 
  • #14
Gokul43201
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vincentchan said:
how do you get the second derivative..... I used conservation of energy.... the rotational KE and translational KE is propotional to [itex] \frac{d \theta}{dt} [/itex] square , PE = [itex] mglsin \theta /2 [/itex], set both side equal....
But that's not conservation of energy !

You set PE = KE. What you should have done is set [itex]\Delta PE = -\Delta KE [/itex], which gives the limit

[tex]\frac{d}{dt}PE = -\frac{d}{dt}KE [/tex]

The Newtonian formulation and the Lagrangian formulation should give the same equation.
 

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