Solving Mechanics Problems with (1)-(4)

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In summary: No, the question was asking for the horizontal and vertical components of the tension forces. But you have correctly solved for those components. In summary, the problem involves a mass attached to two strings that are attached to a rotating apparatus. The horizontal component of the tension forces is equivalent to the centripetal force keeping the object in circular motion, and the vertical net force on the object is equal to the vector sum of the vertical components of the two tension forces and gravity. The equations used to solve for the tension forces are T_{1x}-T_{2x}=mg\cot(\alpha) and T_{1x}+T_{2x}=lm\omega^{2}\sin(\alpha). The solutions for the tensions are T
  • #1
Mentallic
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Homework Statement


http://img21.imageshack.us/img21/3430/mechanics3.jpg

Homework Equations



[tex]w=\frac{d\theta}{dt}[/tex] (1)

[tex]v=rw[/tex] (2)

[tex]F=ma[/tex] (3)

[tex]a=\frac{v^2}{r}[/tex] (4)



The Attempt at a Solution


For (i) to find the horizontal component I used (2) and (4) to get [tex]a=rw^2[/tex] and [tex]r=l.cos\alpha[/tex] hence [tex]a=l.cos\alpha.w^2[/tex]
But I have no idea for the vertical component. I just keep guessing at that one.

I think I'll need the answer to (i) before I can answer (ii)
 
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  • #2
[tex]r=l\sin(\alpha)[/tex] if you draw out the triangle.

In this case of uniform circular motion, the horizontal component of force would be the centripetal force keeping the object in circular motion.

Don't forget that both tensions contribute to the centripetal force, so [tex]T_{1x}+T_{2x}=lm\omega^{2}\sin(\alpha)[/tex].

The vertical net force on P would have a magnitude equal to the vector sum of the vertical components of the two tension forces and gravity. Although not explicitly stated by the problem, the vertical net force is probably 0, so [tex]F_{net,y}=0=T_{1y}-T_{2y}-mg[/tex] (with up as positive).

Btw, most of these look like they should be in the physics section o.o
 
  • #3
zcd said:
[tex]r=l\sin(\alpha)[/tex] if you draw out the triangle.
Ahh yes thanks for correcting me on that.

zcd said:
Don't forget that both tensions contribute to the centripetal force, so [tex]T_{1x}+T_{2x}=lm\omega^{2}\sin(\alpha)[/tex].
So then the force F=ma is equivalent to the tension's of each string holding it in place? Ahh that makes sense :smile:

So [tex]F_x=T_{1x}+T_{2x}=lm\omega^{2}\sin\alpha[/tex]


zcd said:
The vertical net force on P would have a magnitude equal to the vector sum of the vertical components of the two tension forces and gravity.
This too makes sense :smile:

zcd said:
Although not explicitly stated by the problem, the vertical net force is probably 0, so [tex]F_{net,y}=0=T_{1y}-T_{2y}-mg[/tex] (with up as positive).
It's always assumed zero for our course, unless otherwise stated.
Why is it the negative of [tex]T_{2y}[/tex] and not the positive? Maybe because its significance is that it keeps the mass pulled downwards?

zcd said:
Btw, most of these look like they should be in the physics section o.o
Actually I've already learned basics of motion in physics, but we were studying this in maths. We also haven't learned about vectors, so please bare with me here.
 
  • #4
Uhh I'm a little worried what I tried is again, incorrect.

[tex]T_{1y}=l.cos\alpha[/tex] yes?

But also, [tex]T_{2y}=l.cos\alpha[/tex] by the looks of it.

Hence, mg=0 ? which doesn't sound right...
 
  • #5
Mentallic said:
Hence, mg=0 ? which doesn't sound right...

Looks can be deceiving.
 
  • #6
...but in this case they're not. lcos(a) is half the height of the apparatus, not the tension. To use cos(a), you can express Ty in terms of the total tension, T:

T1y=T1cos(a)
T2y=T2cos(a)
 
  • #7
So uh...

[tex]T_{1y}-T_{2y}-mg=cos\alpha(T_1-T_2)-mg=0[/tex]

[tex]cos\alpha(T_1-T_2)=mg[/tex] (1)

[tex]T_{1x}+T_{2x}=sin\alpha(T_1+T_2)=lmw^2sin\alpha[/tex]

[tex]T_1-T_2=lmw^2[/tex] (2)

I'm unsure what to do now. I'm thinking both tensions are supposed to be solved simultaneously, but I wouldn't be able to eliminate both tensions with just 2 equations.
 
  • #8
The problem was asking for forces in vertical and horizontal directions, and that seems to have finished the job. If you want to solve for [tex]|\vec{T}_{1}|[/tex] or [tex]|\vec{T}_{2}|[/tex], note that:
[tex]T_{1y}=T_{1}\sin(\alpha)[/tex]
[tex]T_{1x}=T_{1}\cos(\alpha)[/tex]
[tex]T_{1y}=T_{1x}\tan(\alpha)[/tex]
and proceed with the algebra
 
  • #9
Then if [tex]T_{1y}=T_{1x}tan\alpha , T_{2y}=T{2x}tan\alpha[/tex]

and [tex]T_{1x}+T_{2x}=lm\omega^2sin\alpha[/tex]

then [tex]cot\alpha(T_{1y}+T_{2y})=lm\omega^2sin\alpha[/tex]

So the best I can do with this is: [tex]T_{1y}+T_{2y}=lm\omega^2sin\alpha tan\alpha[/tex]

When you say proceed with the algebra, sure I have no problem doing that for other topics, but in this which is very new to me, all I'm seeing is a big mess where I have a bunch of equations that don't make sense to me... Could you please direct me to which equations I need to be solving for? Maybe [itex]F_x[/itex] with something else?
 
  • #10
Instead of grouping into the same directional component (x in that case), you could try to reduce it to components of the same tension force. For example:
[tex]T_{1x}-T_{2x}=mg\cot(\alpha)[/tex]
[tex]T_{1x}+T_{2x}=lm\omega^{2}\sin(\alpha)[/tex]
system of equations

[tex]T_{1x}=\frac{1}{2}(mg\cot(\alpha)+lm\omega^{2}\sin(\alpha))[/tex]
[tex]T_{1}=\frac{1}{2}(mg\sec(\alpha)+lm\omega^{2}\tan(\alpha))[/tex]
is that what the question is asking for?
 

FAQ: Solving Mechanics Problems with (1)-(4)

What is "Solving Mechanics Problems with (1)-(4)"?

"Solving Mechanics Problems with (1)-(4)" refers to using the four fundamental equations of motion in classical mechanics to solve problems involving the motion of objects.

What are the four fundamental equations of motion?

The four fundamental equations of motion are:

  • 1. v = u + at (equation for finding final velocity)
  • 2. s = ut + 1/2at^2 (equation for finding displacement)
  • 3. v^2 = u^2 + 2as (equation for finding final velocity without time)
  • 4. s = (u + v)t/2 (equation for finding average velocity)

How do I use these equations to solve mechanics problems?

To solve mechanics problems, you will need to identify which equation(s) to use based on the given information and variables involved. You will then need to rearrange the equation(s) to solve for the desired variable.

What are some common mistakes to avoid when solving mechanics problems?

Some common mistakes to avoid when solving mechanics problems include:

  • Incorrectly identifying which equation(s) to use
  • Using the wrong units for the given values
  • Not properly rearranging the equations
  • Forgetting to consider the direction of motion (positive or negative)
  • Using the wrong sign convention for acceleration

Can these equations be used for all types of motion?

These equations can be used for motion with constant acceleration, which includes free-fall, projectile motion, and motion on an inclined plane. They can also be used for motion with variable acceleration, but may require additional equations or techniques to solve for the variables.

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