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Direction of Forces on an Inclined Plane

  • Thread starter tlonster
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  • #1
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Homework Statement



What are the directions of my forces acting on mass m with respect to my "m" frame (see attachment)?

Homework Equations



some trig and transformations maybe...

The Attempt at a Solution



I know I only have tension, weight, and a normal force acting on my particle but I cannot tell their directions. I want to have tension acting in the -Im direction, and by the way I have drawn that frame (Im, Jm, Km) it looks as if the force from gravity will be acting in the -Jm direction. However, I am confusing myself whether or not it actually is. If that's the case then I'm left with the normal force which acts in the Jm and Ii direction, correct? I can then translate Ii into m frame terms if so.
 

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Answers and Replies

  • #2
ehild
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Be a bit more clear: What does the problem say? What are those two frames? How are the axes defined? Your figure is very nice, but I can not understand :redface:

ehild
 
  • #3
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I'm supposed to find the equations of motion in terms of a given function beta and variable theta. the blue frame is inertial (I) and the other two frames are rotated reference frames. The frame I end up in is the green one (M) I already have the system to set up my acceleration a(M/I). Now I need to use F = ma(M/I) to get my equation of motions. I'm given no values for anything, but initial conditions for theta to eventually simulate the system. I'm just confusing myself with the three forces. Theta double dot only shows up in my Km component for acceleration but I feel like the normal force has a component acting in that direction as well. If this is so, I don't know how I'm supposed to simulate this considering my normal force is unknown....along with tension and mass.

Thoughts?
 
  • #4
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Because I only have one component of all these forces acting in my Km direction, can I say that the sum of the forces in that direction is equal to zero? Therefore, my N is just zero. Can a normal force even be zero?
 
  • #5
ehild
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Is it a particle on a slope? What is the "given function beta" ?

The normal force is perpendicular to the surface along the particle is constrained to move. If some force acts on the particle in addition to gravity, and it has a component normal to the slope, the normal force can be zero. The normal force balances the other forces so as the particle stays on the slope, has zero acceleration normal to the slope.

ehild
 
  • #6
ehild
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Does the angle of inclination change with time???

ehild
 
  • #7
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yes, beta is a function of time.

the problem statement is to find the equations of motion for the mass m hanging by a light string on an incline. both beta and theta change with time. i have already computed what the acceleration of the mass is as seen by the inertial frame. now, i need to apply F=ma to figure out the EOM. unfortunately I am stuck. I want my forces to be in the m (green) reference frame. for instance, tension would be -T acting in the direction of Im. I am having trouble with the normal force and the weight. what i think they are is as follows...

W = mg in the Jm direction (unsure)
N = N in the Ip direction which can be translated to m frame components...

Ip = sin(beta)Ii + cos(beta)Jm where Ii = cos(theta)Im - sin(theta)Km

Hence,

N = Nsin(beta)cos(theta)Im - Nsin(beta)sin(theta)Km + cos(beta)Jm

From my acceleration I only have a theta double dot term in my Km component. If I set the above Km component of my normal force to this acceleration component in that direction I have no way to solve.

Could I assume the magnitude of N = mgcos(beta)?

My mass m would drop out when I set F=ma to each other.
 
  • #8
ehild
Homework Helper
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If beta changes with time it is "a moving constraint" problem. You can not say that N=mgcos(theta). The problem is quite complicated, I can not help. It is not "Introductory Physics". Report (click to "Report" to your own post, and ask the Mentor to move it to "Advanced Physics".
Also, try to read http://ruk.usc.edu/bio/udwadia/papers/AMConMotion.pdf

ehild
 

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