How Do Forces Act on an Elevator in Motion and at Rest?

[pinkyjoshi65]

uh..but how can we find the T when we don't know theta or the horizontal component of Tension

 

[-RA-]

good point, you will need to find either Theta or the horizontal component, but you don't have that.

do we know what the height of the rock is compared to when it is at rest?

maybe with that you can use GPE = mgh? to be honest I'm not sure, ill leave this to an expert before i confuse you even more!

 

[pinkyjoshi65]

k..thnkz..

 

[Hootenanny]

the vertical component of the tension is acting in the oppostie direction to the mg of the rock..?

Correct,
the opposing forces are the same, as grvaity is the only force acting on the rock. At any set time the vertical componet of the tension of the string = mg.[/quote]
Again, correct

uh..but how can we find the T when we don't know theta or the horizontal component of Tension

Also correct..

good point, you will need to find either Theta or the horizontal component, but you don't have that.

maybe with that you can use GPE = mgh? to be honest I'm not sure, ill leave this to an expert before i confuse you even more!

Sounds like a plan to me.

So from the above statements we know that the vertical component of the tension is equal to the weight, in maths that would be;

T\cos\theta = mg

Do you follow? Assuming you do the we also know that the vertical side of your right triangle can be written as;

R\cos\theta = R-h

Where R is the length of your string and h is the vertical height of the stone above our zero potential (the bottom of the circular path in the previous questions [point B]). If you can't see where this came from check your diagram.

Once you understand where that came from we need to move onto calculating the tension, but for that we need to know the velocity. We need to be careful at this point, the circular path of the rock traces out the base of the cone (check the link I provided) and therefore the radius of the circular path (lets call it r is not the length of the string R, it will actually be the length of the base of the triangle i.e;

r = R\sin\theta

Do you follow?

Therefore, for the net horizontal force for the circular motion can be written as;

F_x = \frac{mv^2}{r} = \frac{mv^2}{R\sin\theta}

Do you follow?

I know this is a heavy post, so we'll leave it there for now and we'll pick it up once you've digested all of it.

 

[pinkyjoshi65]

ahan..

 

[Hootenanny]

Now it's your turn to do some work. Can you write a conservation of energy equation for the potential and kinetic energy of the particle?

 

[pinkyjoshi65]

E_t= E_k+E_p= 0.5mv^2+ mgh

 

[Hootenanny]

E_t= E_k+E_p= 0.5mv^2+ mgh

Now, if we assume that the total energy must be the same as that for the vertical path we can write;

mg(2R) =\frac{1}{2}mv^2 + mgh

4gR = v^2 + 2gh

Do you follow?

 

[pinkyjoshi65]

yes..

 

[Hootenanny]

Good, so can you now take the final equation and re-write it as a function of \theta instead of h?

 

[pinkyjoshi65]

yes..4gR=v^2+2g*R(1-costhetha)..?

 

[Hootenanny]

yes..4gR=v^2+2g*R(1-costhetha)..?

Correct! So know we can re-write the following equation;

F_x = \frac{mv^2}{r} = \frac{mv^2}{R\sin\theta}

as

F_x = \frac{m\left[ 4gR-2gR(1-\cos\theta)\right]}{R\sin\theta}

We can also re-write F_x as;

F_x = T\sin\theta

Hence, we can re-write our main equation as

T\sin\theta = \frac{m\left[ 4gR-2gR(1-\cos\theta)\right]}{R\sin\theta}

We also know that;

T\cos\theta = mg \Leftrightarrow T = \frac{mg}{\cos\theta}

And we can finally re-write our expression;

\frac{mg\sin\theta}{\cos\theta} = \frac{m\left[ 4gR-2gR(1-\cos\theta)\right]}{R\sin\theta}

Do you follow?

 

[pinkyjoshi65]

yes..

 

[Hootenanny]

yes..

So now it's your turn, solve for \cos\theta...

 

[pinkyjoshi65]

ok i got a quadratic equation---8.35Cos^2thetha-14.7Costheta+7.35=0..When i used the Quadratic equation, i got a negative root...:S

 

[Hootenanny]

ok i got a quadratic equation---8.35Cos^2thetha-14.7Costheta+7.35=0..When i used the Quadratic equation, i got a negative root...:S

Well I'll start you off and we'll see how you go; let's start by cancelling the mg's and the R's;

\frac{\cancel{mg}\sin\theta}{\cos\theta} = \frac{\cancel{m}\left[ 4\cancel{g}\cancel{R}-2\cancel{g}\cancel{R}(1-\cos\theta)\right]}{\cancel{R}\sin\theta}

Now that leaves us with;

\frac{\sin\theta}{\cos\theta} = \frac{4-2(1-\cos\theta)}{\sin\theta}

Expanding the numerator;

\frac{\sin\theta}{\cos\theta} = \frac{4-2+2\cos\theta}{\sin\theta}

Getting rid of the fraction on the LHS;

1 = \frac{2\cos\theta+2\cos^2\theta}{\sin^2\theta} = \frac{2\cos\theta+2\cos^2\theta}{1-\cos^2\theta}

Moving the denominator to the LHS;

1-\cos^2\theta = 2\cos\theta+2\cos^2\theta

Collecting terms;

3\cos^2\theta +2\cos\theta - 1 = 0

Can you go from here?

 

[pinkyjoshi65]

by solving this i get the angle as 78 degrees, but the answer is 70 degrees.

 

[pinkyjoshi65]

ohk..never mind i got it..!..thanks..:)