Simple Tension Problem: Free Body Diagram and Calculating Normal Force

[Kant Destroyer]

1. Homework Statement [/b]
A block with mass M = 5kg sits at rest on a frictionless incline. The mass is connected to the wall by a string with linear density μ = 5.0 g/m. The incline is fristionless, with angle Θ = 30°. Let the positive x-direction point up along the incline, and let the origin (x=0) be at the end of the string attached to the mass. HERE IS A DIAGRAM OF WHAT THE PROBLEM LOOKS LIKE: http://i.imgur.com/puAhYlT.png

Draw a free body diagram of the block. What is the normal force on the block?

Homework Equations

This is probably where I'm lacking. I'm not sure if there is an equation that will give me a way of solving for T or not, but other than that there are no real equations necessary.

The Attempt at a Solution

I'm having no real problems with the free body diagram. It's simple enough. The issue I'm having is that I don't know how to calculate the Normal force on the block due to the incline. I calculate that:

N$\ast$sin(60) + T$\ast$sin(30) = mg
N = $\frac{(mg - 1/2T)}{sin(60)}$
N = $\frac{(2mg - T)}{\sqrt{3}}$

But now the only way I can see myself solving for N is with the value of T, which I can't seem to figure out. Am I just missing a valuable equation?

 

[Doc Al]

But now the only way I can see myself solving for N is with the value of T, which I can't seem to figure out. Am I just missing a valuable equation?

You're making life hard for yourself by insisting on using horizontal and vertical components. You can certainly solve the problem that way, but you'd better be careful with signs.

Much easier: Use components parallel and perpendicular to the incline surface. Then your equations will be simpler.

 

[Chestermiller]

Try resolving the forces into components parallel and perpendicular to the incline.

Oops. Doc Al scooped me.

Chet

 

[Doc Al]

I see that your diagram already has x and y defined as parallel and perpendicular to the incline. So use that!

 

[haruspex]

There are two ways forward:
You can get a second equation by looking at vertical forces. T and N will both feature again.
Alternatively, instead of resolving horizontally and vertically, resolve parallel to and orthogonal to the slope. Now one equation involves T while the other involves N.

Btw, I assume there's more to the question since we have not used the string density yet. Strictly speaking, the string density does have a tiny effect on the normal force, but I hope we're supposed to ignore that. The string will sag under its own weight, so that where it meets the block it will not be quite parallel to the ramp. This will slightly increase the normal force. Finding that angle would involve the equation for a catenary.

 

[Doc Al]

Btw, I assume there's more to the question since we have not used the string density yet.

Good point!

 

[Kant Destroyer]

I see that your diagram already has x and y defined as parallel and perpendicular to the incline. So use that!

I'm not sure exactly what you mean. I had a feeling that I was doing it wrong because of the different orientation of x and y, but I just went with what I'm used to. Are you suggesting that I just set N equal to the y component of the weight?

 

[haruspex]

I'm not sure exactly what you mean. I had a feeling that I was doing it wrong because of the different orientation of x and y, but I just went with what I'm used to. Are you suggesting that I just set N equal to the y component of the weight?

Yes.

 

[Doc Al]

I'm not sure exactly what you mean. I had a feeling that I was doing it wrong because of the different orientation of x and y, but I just went with what I'm used to. Are you suggesting that I just set N equal to the y component of the weight?

Yes.

That comes from applying the equilibrium condition (that the net force = 0) in the y direction. The only forces in that direction are the normal force (up) and the y component of the weight (down).

 

[Kant Destroyer]

Yes.

That comes from applying the equilibrium condition (that the net force = 0) in the y direction. The only forces in that direction are the normal force (up) and the y component of the weight (down).

Thank you. I'm working it out right now. It's a 45-45-90 triangle, so would the Tension be equal to the normal force in magnitude?

 

[Doc Al]

It's a 45-45-90 triangle, so would the Tension be equal to the normal force in magnitude?

Where did you get 45 degrees? The angle of the incline is 30°.

 

[Doc Al]

To find the tension, set up an equation forces in the x direction.

 

[Kant Destroyer]

Where did you get 45 degrees? The angle of the incline is 30°.

I think I'm just confused about the angles. I'm not sure how to set up the right triangle and solve for it's different vector parts with this axis system because I can't seem to figure out what the triangle would look like. Is it a 30-60-90 triangle with the long leg being the y component and the short leg being the x component? In that case the normal force would just be equal to 1/2*weight*sqrt(3) and tension would be equal to 1/2*weight.

 

[Doc Al]

I think I'm just confused about the angles. I'm not sure how to set up the right triangle and solve for it's different vector parts with this axis system because I can't seem to figure out what the triangle would look like. Is it a 30-60-90 triangle with the long leg being the y component and the short leg being the x component? In that case the normal force would just be equal to 1/2*weight*sqrt(3) and tension would be equal to 1/2*weight.

Yes, that is correct.

I suggest that you learn to think in terms of vector components. Analyzing y components gives you: N = mg cos30° = mg(√3)/2. Analyzing x components gives you: T = mg sin30° = mg/2.

You might want to review incline planes and vectors: Inclined Planes