Calculating tension for mass on incline slope

eazyduzit

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Could anyone help me answer this question, I always seem to end up with 2 simultaneous equations which I cant seem to solve.

PYJkF0g.jpg
 
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Hi eazyduzit,

My advice for solving this problem would be

1) Set up a Cartesian coordinate system that is aligned with the inclined plane. I.e. the \(\displaystyle x\) coordinate varies in the direction along (parallel to) the surface of the plane, and the \(\displaystyle y\) coordinate varies in the direction perpendicular to the surface of the plane. You have to pick a sign convention too. Let's say that the \(\displaystyle +x\) direction is the direction moving up the incline, while sliding down the incline is decreasing \(\displaystyle x\). Let's say that the \(\displaystyle +y\) direction is moving away from the surface of the incline, while moving towards it decreases \(\displaystyle y\). Orienting the coordinate axes to line up with the incline is simply more convenient in this situation than choosing them to be horizontal and vertical (relative to the page).

2) Draw a free body diagram of the box. This is a diagram in which you consider the box in isolation, drawing only it, and the forces that act on it. That makes it much easier to see what is affecting the state of motion of the box, without being confused by extraneous details in the surroundings. In order to do this, you have to make a tally of all the forces that are acting on the box. As far as I can tell, these are:


  • the box's weight (call this \(\displaystyle w\)). I.e. the force due to gravity on the box. As always, \(\displaystyle \vec{w} = m\vec{g}\), where m = 2 kg is the mass of the box, and \(\displaystyle \vec{g}\) = 9.81 N/kg, downward, is the acceleration due to gravity (on Earth's surface). The weight force vector acts vertically downward, which means that you will need to resolve it (i.e. decompose it) into \(\displaystyle x\) and \(\displaystyle y\) components \(\displaystyle w_x\) and \(\displaystyle w_y\)



  • The force due to the tension in the rope (call this \(\displaystyle T\)). The tension force vector \(\displaystyle \vec{T}\) acts at an angle of \(\displaystyle \alpha\) above the surface of the plane, which means that you will need to resolve it into components \(\displaystyle T_x\) and \(\displaystyle T_y\).



  • The force due to friction with the surface of the plane (call this \(\displaystyle f\)). This force acts directly in the \(\displaystyle x\) direction. If we assume that, in the absence of any forces other than gravity, the box would want to move in the \(\displaystyle -x\) direction (to slide down the plane), then the frictional force acts purely in the \(\displaystyle +x\) direction to oppose this motion. The magnitude of the friction force is given by \(\displaystyle f = \mu N\), where \(\displaystyle N\) is the normal force acting on the box (see below). The coefficient of friction (of static friction in this case), \(\displaystyle \mu\) is stated to be 1/3 for these two surfaces (box and plane).



  • The normal force (\(\displaystyle N \)) on the box from the surface of the plane. This is a reaction force (the surface of the plane pushes on the box). The word "normal" means "perpendicular" in this context, and this is the force on the box that acts purely perpendicular to the surface of the plane i.e. purely in the \(\displaystyle +y\) direction.

3) The statement that the box is in equilibrium is a statement that there is no net force acting on it. In other words, all the above listed force vectors, when summed up as vectors, add up to zero resultant force:

\(\displaystyle \displaystyle \sum \vec{F} = \vec{0} \)

We can break this sum up into two separate sums of the components of the forces in each of the \(\displaystyle x\) and \(\displaystyle y\) directions (where these components have signs):


\(\displaystyle \displaystyle \sum F_x = w_x + T_x + f = 0 \)

\(\displaystyle \displaystyle \sum F_y = w_y + T_y + N = 0 \)

Can you take it from here? You need to actually compute each of these force components in terms of known quantities.

Note that the procedure I've outlined above is the approach to take in solving any statics problem, not just this one. :)
 
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Could anyone help me answer this question [link to prohibited site deleted] I always seem to end up with 2 simultaneous equations which I cant seem to solve.

In order to give specific help, we need to know whether your equations are wrong, or you are wrong about not being able to solve them.

Can you at least show us the equations you got (along with any variable definitions that are needed in order to check them)?
 
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So I've had another crack at it, I started with a body diagram and started simplifying here taking Fr as frictional force, Nr and the normal reaction force and T as obviously tension.
iNWHFT4.jpg

I then resolved the vertical and horizontal components and since the body is in equilibrium I derived the two equations on the right.
gDQQExj.jpg 7YdX77I.jpg

I tried to solve them and came up with this.
WLD1O9t.jpg

Anyone able to shed any light on it?

Here's a more detailed version of my working out.
mwBJ0Iv.jpg
 
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So I've had another crack at it, I started with a body diagram and started simplifying here [link to prohibited site deleted] taking Fr as frictional force, Nr and the normal reaction force and T as obviously tension.

I then simplified the vertical and horizontal components to get this [link to prohibited site deleted] and since the body is in equilibrium I derived these two equations [link to prohibited site deleted]

I tried to solve them and came up with this [link to prohibited site deleted]

Anyone able to shed any light on it?

Here's a bit more detail of my working out [link to prohibited site deleted]
Please reply with images that have been uploaded to this site, or with your typed-out text. Thank you! ;)
 
I can't seem to upload them directly for some reason.
Check the file sizes. They may be too large. (Or post links to an image site that does not post other stuff that isn't good for the younger portions of this site's clients.)
 
So I've had another crack at it, I started with a body diagram and started simplifying here taking Fr as frictional force, Nr and the normal reaction force and T as obviously tension. QUOTE]

it seems you got a direction wrong.

in direction of the plane you have 3 forces,
wt, friction, and rope. You know the first 2.
wtsin30=1/3wtcos30+Fropecos&

& is angle of rope with inclined plane.
 
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