Deriving the Minimum Tension for a Particle on an Inclined Plane

[Jadenag]

Okay on question two i can only get as far as drawing fbd for the particle. I realize I need to get sin/cos for tan but don't really know how to get there.

question three i have no idea. I don't understand. if the string is at an angle then does that mean tension has two components? If so then the vertical component the tsinx component what happens to this?

 

[LawrenceC]

Here is a hint on Q2:

Draw a free body diagram and note that if the angle of the rope is not zero, the friction force changes to mgu-Fsin(theta). Fsin(theta) is the upward component of the pulling force. Now write an equation that represents the free body diagram. Since a minimum is sought, that implies some calculus is needed. In my above equation, m is mass, g is gravity acceleration, u is friction coefficient, and F is the tension in the rope.

 

[Jadenag]

Here is a hint on Q2:

Draw a free body diagram and note that if the angle of the rope is not zero, the friction force changes to mgu-Fsin(theta). Fsin(theta) is the upward component of the pulling force. Now write an equation that represents the free body diagram. Since a minimum is sought, that implies some calculus is needed. In my above equation, m is mass, g is gravity acceleration, u is friction coefficient, and F is the tension in the rope.

Im sorry I don't understand

I have made two equations.
TSinθ + R= mg (i)
TCosθ = μ₂(R)

I don't know where to go from here?

 

[LawrenceC]

You are heading in the right direction. In your second equation substitute the function for R that you have in the first equation. Once you've done this you have a function that represents the tension in terms of theta and friction coefficient because R has been eliminated. Put T on left side of equation and all the other parts to right of equals mark.

Now, since the question asks for the minimum, you find it by taking the derivative with respect to theta. Set the derivative to zero and you'll see that much of what remains can be removed because of the 0 on the right hand side of equation. Once you've done this, you'll end up with what is requested.

To be complete, you must prove that what you have is a minimum. That necessitates taking the second derivative of the original function for tension and showing that it is positive. This would prove that you have a minimum.

I am traveling today so this is all the help I can supply.

 

[Jadenag]

You are heading in the right direction. In your second equation substitute the function for R that you have in the first equation. Once you've done this you have a function that represents the tension in terms of theta and friction coefficient because R has been eliminated. Put T on left side of equation and all the other parts to right of equals mark.

Now, since the question asks for the minimum, you find it by taking the derivative with respect to theta. Set the derivative to zero and you'll see that much of what remains can be removed because of the 0 on the right hand side of equation. Once you've done this, you'll end up with what is requested.

To be complete, you must prove that what you have is a minimum. That necessitates taking the second derivative of the original function for tension and showing that it is positive. This would prove that you have a minimum.

I am traveling today so this is all the help I can supply.

Got it! Thanks!