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Mass on an Incline

  • Thread starter rpg711
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  • #1
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Homework Statement


A block with mass m1 = 9.2 kg is on an incline with an angle θ = 25.0° with respect to the horizontal. For the first question there is no friction between the incline and the block.

The coefficient of static friction was deterimined to be μ = .121

The premise of this question is that there was a spring holding the box up(the spring only exerting force in the X direction), but this spring was replaced by a STRING that is parallel to the horizontal, thus making a 25° angle with the incline.

Word for word, it is:
The spring is replaced with a massless rope that pulls horizontally to prevent the block from moving. What is the tension in the rope?

Homework Equations


Fgx = mgsin(theta)
Fgy = mgcos(theta)

Fty = Ftsin(theta)
Ftx = Ftcos(theta)

Ff = μN

N = mg + Fty

The Attempt at a Solution


This problem has been frustrating me a lot, I've probably been stuck for hours on this one part on my already done homework... I just want to know how it's done because I'm definitely not going to be able to get credit at this point...

I've figured that Ff = 9.886 + μ(Fty) and so the force of friction is dependent on the tension force in the Y direction. Also, the tension force is dependent on the force of friction because Ftx = Fgx - Ff. It seems impossible for me because every time I try to setup a systems of equations, I end up with 0 = 0, which doesn't really help me. I've tried dozens of approaches, this one seems to be the closest because it factors in the tension in the Y direction into the normal force, which incorporates into the frictional force, but I can't figure it out.

A solution would be greatly appreciated.
 
Last edited:

Answers and Replies

  • #2
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Have you tried to draw free body diagrams of the block in both scenarios? A general diagram of the whole setup will assist you for this purpose (and in general).
 
  • #3
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Have you tried to draw free body diagrams of the block in both scenarios? A general diagram of the whole setup will assist you for this purpose (and in general).
Yep, and determined that the only two forces acting UP the incline is friction and tension force(the parallel X component), and also that there is a Y component of tension perpendicular to the slope that would increase the normal force, which would increase the friction, which would reduce the tension. It's just a confusing loop and I can't seem to setup the system of equations(which I presume is the only way this can be solved)

BTW I corrected myself, it's coefficient of static friction, not kinetic.
 
  • #4
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I am not sure if I am picturing this correctly, so I drew how I understand the situation. Correct me if that is wrong.

If not, then there is no component of tension force perpendicular to the incline, at the box. Notice that I have drawn all the forces on the box.
 
  • #5
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Nope, the string is connected in such a way that it is parallel to the HORIZONTAL, so it makes an angle of 25 degrees with the incline.

So there is a component parallel to the incline(Tcos(25)) and a component perpendicular to the incline that has to be factored into Friction/Normal force calculation (Tsin(25))
 
  • #6
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A clarification: In that picture, the dotted line is the string, so it is parallel to the horizontal.

Can you draw the correct setup?
 
  • #7
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Here's the correct setup
 

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  • #8
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Ah, I see. It seems you have solved the problem in your first post. Below I have quoted it indicating the substitutions you need to make (replace darker shade by lighter shade of same color). I do not get a 0 = 0 so I'm guessing you made some algebraic error. Do it again and see if you still get such discrepancy and post it here if you do.

Homework Equations


Fgx = mgsin(theta)
Fgy = mgcos(theta)

Fty = Ftsin(theta)
Ftx = Ftcos(theta)

Ff = μN

N = mg + Fty

The Attempt at a Solution


I've figured that Ff = 9.886 + μ(Fty) and so the force of friction is dependent on the tension force in the Y direction. Also, the tension force is dependent on the force of friction because Ftx = Fgx - Ff.
 

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