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SONG FU QIAN September 4, 2011 at 1:58 am

ThE sUrFaCe Of ThE cUbE aNd ThE wHaT eQuAl To ItS lEnGtH?

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Neal September 5, 2011 at 1:38 pm

A cube has six surfaces of equal area. So the surface area is 6 x L^2.

You could take that and say that you could divide the surface area by six and then take the square root to get the length.

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Daniel King September 11, 2011 at 9:43 pm

still don’t understand #53 & 67. For some reason I don’t get the answers that is in the book. Help please.

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Neal September 12, 2011 at 11:49 pm

You need trigonometric relations to use the angle and get the right component for acceleration. Then you use the two kinematic equations

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Minji September 14, 2011 at 9:56 am

For the Vector Addition Force lab is it reasonable to take F1 = kg*a and then take that to get the vector F (cos theta) i, F (sin theta) j to get the F1x and F1y and then add them up to get the resultant vector (Rx = F1x + F2x + F3x) (Rx= F1y + F2y + F3y) ?

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Neal September 14, 2011 at 11:03 am

Hey Minji, yes, that works.

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Minji September 14, 2011 at 10:09 am

Also, do you have to take F=Sqrt. Fy2 + Fy2 first to get the F first and when do you take the (tan theta) = Fy / Fx?

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Neal September 14, 2011 at 11:05 am

If you started off with the components, you would use the Pythagorean Theorem to get the magnitude of the resultant vector. You would use tangent if you had the two sides of a right triangle not including the hypotenuse since sine and cosine require the hypotenuse.

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Minji October 18, 2011 at 2:40 pm

Chapter 7 # 24.
Py = 4 m, 20 kg
Px = 2 m, 10 kg
How would you factor in the torque?
We are stuck…. Daniel and I derived the T net = 300 N m.
Is that right?
We are confident in our answer but unsure?

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steven Franklin November 15, 2011 at 9:05 am

I have 2 ropes supporting a weight of 700N. 1 rope has an angle of 60 degress down from the horizontal and the other has an angle of 40 degrees doewn from the horizontal. What is the tension in each of the supporting ropes?

I got an answer but don’t like it because the tension I have along the rope dropped at 40 degrees is greater than the other.

I need help. I think I must have gotten the sign wrong somewhere.

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Neal November 15, 2011 at 5:36 pm

Hello Steven,
So what happens in that situation is the horizontal components of tension in the two ropes need to be equal, otherwise the weight would be accelerating to either the left or right. The vertical components of the tension add up to 700 N, I believe. Depends on the set up of the problem. Are you given the tension in one rope? That is how similar problems I have seen are set up.

Once you have the horizontal components of the tension in each rope, you should be able to use trigonometry to get the tension forces in the two ropes.

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Minji February 10, 2012 at 9:32 am

Can you please post the hw for Ch. 18.
Thanks.

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Neal February 10, 2012 at 2:15 pm
Mike March 13, 2012 at 4:28 pm

Could you post the homework assignment for chapter 21.
Thanks

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Neal March 13, 2012 at 7:01 pm

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