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

[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..:)