Finding Parallel/Perpendicular Forces & Normal Force

[jubbly]

Hey guys, I was wondering if you can help tell me if I am doing this right for finding the parallel and perpendicular force.

To find the parallel force would I have to take: gravity*sin(theta) or would I have to take the weight? Same goes with the perpendicular force, but using cosine.

Also to find the normal force would I take the weight-tension*sin(theta)?

Thanks for your help.

 

[Lancelot59]

Force of what? Do you have a specific situation? It's not so much parallel and perpendicular. What you're finding there are the components of that force in the x, and y axes.

Now your normal force for example is applied by a surface in response to the force applied by an object resting on it. Remember, for every action there is an equal and opposite reaction. If you have a box sitting on a table, gravity pulls down on the box. This exerts a downward force on the table, which in turn exerts a force of equal magnitude back in the opposite direction. If you were pushing down on said box, the normal force would then be equal to the force of gravity, plus the force you were applying to the box.

 

[jubbly]

I'm just analyzing forces with numbers given to me.
The weight, tension, and the angle is the only given I have.

 

[Lancelot59]

Well posting the actual question would be helpful.

 

[jubbly]

It's just a table, titled analysis of forces.
I am supposed to use the weight, angle, and tension given to find the perpendicular, parallel and normal forces.

The weight I have is -10 at an angle of 15 degrees and the tension is 1.5.
Sorry if I can't be anymore clearer than this, not very good at physics.

 

[Lancelot59]

I see then. Okay. You can solve pretty much any physics problem with three steps:

1. Free body diagram
2. Net force equation
3. Solve it

Do you know what a free body diagram is, and how to properly make one?

 

[jubbly]

I don't know about properly making one, but I do know what it is.

So if I drew one at an angel of 0 it would look like a straight line with an object in the middle?
Then I would draw a arrow below the object labeling that Fg and on top of the object would be FN. Is there anything else I am missing?

 

[Lancelot59]

Well you're supposed to draw them originating from the centre of the object. You have the correct idea I think. Fg points down, Fn points up, and Ft points in whatever direction Ft points in. Your 0 angle should be parallel with the positive X axis (pointing to the right).

Looking at that description I'd have to say the object is on a 10 degree incline with a rope pulling up the slope.

Now that you have your forces figured out, it's time to put them into a net force equation.

Now your box isn't accelerating, so using Fnet=ma your net force is equal to zero. Therefore Fnet=Ft+Fg+Fn=0.

Now you need to break it up into your X and Y components. Let's rotate our reference frame to make the incline zero degrees. So now your Fg is acting at a 15 degree angle relative to your reference frame.

Looking at the components now, you have in your X axis:
Fnet=Fg(x)+Ft=0
and in your Y axis
Fnet=Fg(y)+Fn=0

Now you can rewrite these as:
Fg(x)=Ft
Fg(y)=Fn
Do you see why you can do this?

 

[jubbly]

Oh, now I understand. Thanks for clearing things up for me!