# Forces in Equilibrium (Vectors) II

• Sep 28th 2008, 04:42 PM
Oblivionwarrior
Forces in Equilibrium (Vectors) II
Here is another question I am suck on, it seems simple but I cant get it.

http://i33.tinypic.com/2hycjdx.jpg

Maybe i am drawing the free body diagram wrong. Does the tension in the rope still need to be include since its the same on both sides? (since there is one rope). Thanks again!
• Sep 28th 2008, 04:57 PM
Chris L T521
Quote:

Originally Posted by Oblivionwarrior
Here is another question I am suck on, it seems simple but I cant get it.

http://i34.tinypic.com/rjlg5y.jpg

Maybe i am drawing the free body diagram wrong. Does the tension in the rope still need to be include since its the same on both sides? (since there is one rope). Thanks again!

The tension in the rope must be the same at the end of both boxes.

Let the blue box have a mass of $\displaystyle m_1$ and let the orange box have a mass of $\displaystyle m_2$

Doing a FB analysis on the blue box [the x direction is what matters] , we see that

$\displaystyle \sum F_x=0$:

$\displaystyle -m_1g\cos\alpha+T=0\implies T=m_1g\cos\alpha$

Doing a FB analysis on the orange box [the x direction is what matters] , we see that

$\displaystyle \sum F_x=0$:

$\displaystyle m_2g\cos(30^{\circ})-T=0\implies T=m_2g\cos(30)$

Therefore, since the tensions are the same, we see that $\displaystyle m_2g\cos(30)=m_1g\cos(66)$

Can you take it from here?

--Chris
• Sep 28th 2008, 06:01 PM
Oblivionwarrior
Ooh that makes sense now, I was adding the normal force into the mix before. It was just making it complicated. So you don't need the normal force?
• Sep 28th 2008, 06:06 PM
o_O
Sorry, shouldn't the force due to gravity along the incline be $\displaystyle mg\sin \theta$? Otherwise it'd be the rest of the analysis follows: $\displaystyle m_{1} \sin 66^{\circ} = m_{2} \sin 30^{\circ}$

unless I've thoroughly confused myself...
• Sep 28th 2008, 06:07 PM
Chris L T521
Quote:

Originally Posted by o_O
Sorry, shouldn't the force due to gravity along the incline be $\displaystyle mg\sin \theta$? Otherwise it'd be the rest of the analysis follows: $\displaystyle m_{1} \sin 66^{\circ} = m_{2} \sin 30^{\circ}$

unless I've thoroughly confused myself...

I'm the one that confused myself! (Rofl)

--Chris
• Sep 28th 2008, 06:11 PM
Chris L T521
Quote:

Originally Posted by Oblivionwarrior
Ooh that makes sense now, I was adding the normal force into the mix before. It was just making it complicated. So you don't need the normal force?

No. It just tells us that $\displaystyle N_1=m_1g$ and $\displaystyle N_2=m_2g$. It wouldn't tell us much, because the masses aren't known. Since their is tension in the string and they are inclined at different angles, it would have been best to look at the forces along the incline, rather than perpendicular to the incline.

--Chris
• Sep 29th 2008, 04:04 PM
Oblivionwarrior
Are you sure its http://www.mathhelpforum.com/math-he...d52bb865-1.gif? I got the same thing but with Cos.
• Sep 29th 2008, 05:14 PM
skeeter
sine, not cosine.
• Sep 30th 2008, 07:05 PM
Oblivionwarrior
Would the ratio be 1.83:1?
• Sep 30th 2008, 08:29 PM
Chris L T521
Quote:

Originally Posted by Oblivionwarrior
Would the ratio be 1.83:1?

Yes! (Yes)

$\displaystyle 1.83:1$ would signify that $\displaystyle m_2$ is $\displaystyle 1.83$ times larger than $\displaystyle m_1$.

--Chris
• Oct 1st 2008, 05:55 PM
Oblivionwarrior
Quote:

Originally Posted by Chris L T521
Yes! (Yes)

$\displaystyle 1.83:1$ would signify that $\displaystyle m_2$ is $\displaystyle 1.83$ times larger than $\displaystyle m_1$.

--Chris

I put 1.83:1 as the answer and it was wrong, did I screw up the ratio? Should It be 0.55:1 or 1:0.55 or 1:1.83 (Crying)
• Oct 2nd 2008, 01:47 PM
skeeter
an "online" response, no doubt. the ratio for $\displaystyle \frac{M}{m}$ is correct ... how many decimal places do you need?

$\displaystyle \frac{\sin(66)}{\sin(30)} \approx 1.82709091529$
• Oct 2nd 2008, 02:50 PM
Oblivionwarrior
Quote:

Originally Posted by skeeter
an "online" response, no doubt. the ratio for $\displaystyle \frac{M}{m}$ is correct ... how many decimal places do you need?

$\displaystyle \frac{\sin(66)}{\sin(30)} \approx 1.82709091529$

I got it to the right amount of decimals, but after reading the questions again " Determine the ratio of the mass of the right box to the mass of the left box" would the ratio be reversed? 0.55:1?
• Oct 2nd 2008, 03:28 PM
skeeter
left side ... right side

$\displaystyle mg\sin(66) = Mg\sin(30)$

$\displaystyle \frac{\sin(66)}{\sin(30)} = \frac{M}{m}$

what else can I say?
• Oct 2nd 2008, 05:43 PM
Oblivionwarrior
Quote:

Originally Posted by skeeter
left side ... right side

$\displaystyle mg\sin(66) = Mg\sin(30)$

$\displaystyle \frac{\sin(66)}{\sin(30)} = \frac{M}{m}$

what else can I say?

Hmmm, maybe I should just enter just the 1.827 instead of 1.827:1?