Finding the Tension in terms of weight and the angle of incline

AI Thread Summary
The discussion focuses on calculating the tension in a rope connecting two blocks on a frictionless incline. The user attempts to derive the tension using the components of weight and the equation ƩF=ma, but questions the validity of their approach since the blocks are stationary. It is clarified that the blocks are held in place by tension, not by friction, and that the tension depends solely on the mass of one block. The correct understanding is that since there is no acceleration, the net force in the x-direction must equal zero. The tension in the rope is determined by the gravitational force acting on the block down the slope.
Yosty22
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Homework Statement



Two blocks, each with weight ω, are held in place by a frictionless incline.
In terms of ω and the angle θ of the incline, calculate the tension of the rope connecting the two blocks.

Homework Equations



ƩF=ma

The Attempt at a Solution



Since it is on a tilted incline, with angle θ, I made the x and y-axis tilted for the free body diagram. This means that ω, the weight, can be broken down into components ωx, which is parallel to the tension, and ωy, perpendicular to tension. In this case, I assumed that all forces acting in the x-direction (parallel to the tension) were ωx and T. I then used the equation ƩF=max and plugged in values and solved for T. First, I put "ωsinθ" in place of ωx, making the equation ωsinθ+T=ma. Solving for T, I get T=ma-ωsinθ. Is this correct? It looks a little odd to me.
 
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I think something is missing in the statement of the problem. How can a frictionless incline hold anything in place?
 
tms said:
I think something is missing in the statement of the problem. How can a frictionless incline hold anything in place?
No, they are held in place by tension.
 
Yosty22 said:

Homework Statement


Since it is on a tilted incline, with angle θ, I made the x and y-axis tilted for the free body diagram. This means that ω, the weight, can be broken down into components ωx, which is parallel to the tension, and ωy, perpendicular to tension. In this case, I assumed that all forces acting in the x-direction (parallel to the tension) were ωx and T. I then used the equation ƩF=max and plugged in values and solved for T. First, I put "ωsinθ" in place of ωx, making the equation ωsinθ+T=ma. Solving for T, I get T=ma-ωsinθ. Is this correct? It looks a little odd to me.
Since the blocks are held in place there is no acceleration so ma=0⇔ƩFx=0. And remember, there are two blocks.
 
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lep11 said:
No, they are held in place by tension.
I can't picture the setup.
 
Sorry guys, this is the setup. I have looked at it a few more times and still get the same thing. Am I doing something wrong?
 

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Since the blocks are not moving, what is the acceleration?
 
Your answer above is wrong.

It may help to tell you that the tension in that rope would only depend on the mass of B (and not A). Remember the tension in string is equal to the force with which gravity is pulling the mass down the slope.
 
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