Recent content by odie5533

When I listed 600g, I meant the g as in 9.81 not grams or kilograms. When I computed the answers, I multiplied things by 9.81 wherever g is. I use g so that I get a more accurate answer. Am I using all the right forces though? Tension in cable pulling down and to the left, normal force from...

Thanks for the help on the problem, and the explanation of why it's possible, which really helped me understand this problem better. Thanks again!

Homework Statement (a) Compute the torque developed by an industrial motor whose output is 225kW at an angular speed of 4260 RPM. (b) A drum with negligible mass, 0.660 m in diameter, is attached to the motor shaft and the power output of the motor is used to raise the weight hanging from the...

Are all the forces right though? Am I missing any? Here's the third equation: The angle between the strut and the tension from the cable is 16 degress: 180 - 30 - (180 - 46) = 16 \sum\tau = Tlsin16 - \frac{1}{2}l650g(cos46) - 8900l(cos46) = 0 \frac{P_{x}}{cos30}lsin16 = 325lg(cos46) +...

Homework Statement http://img404.imageshack.us/img404/248/sdaft7.png The Attempt at a Solution The arrows on the diagram are drawn in, as are the labels (except for the angles). \sum F_{x} = P_{x} - Tcos30 = 0 T = \frac{P_{x}}{cos30} \sum F_{y} = N_{p} - 8900N - 650g - Tsin30 = 0...

I copied the question word for word. It took me a long time to figure out what it was asking, and I agree it makes more sense for it to be and. Ah well, my professor is notorious for typos and unanswerable questions, so this must just be another one. Thanks for the help.

Homework Statement A 20g cricket leaps through a 2.55 m horizontal distance. The initial velocity of the cricket makes an angle of 30 degress with the horizontal direction. If it takes the cricket 0.019s to leave the ground, what is the magnitude of the impulse of the total force on the...

Ah, I see where I went wrong. I let the cylinder start at 5.236 and go to 1.745 instead of start at 0 and go to 1.745. The very last line should read: \tau = (0.95506)\frac{1.745 - 0}{0.20s} = 8.3 Is this better?: I_{d}\omega_{1} = I_{d}\omega_{2} + I_{c}\omega_{2} \omega_{2} =...

I got this one wrong. The correct answer is 8.3 Nm. I was wondering, could anyone explain to me where I messed up?

I asked my physics prof before class and he said to use the escape velocity in the kinetic energy equation: V = \sqrt{\frac{2GM_{J}}{R_{J}}} K = \frac{1}{2}MV^2 = \frac{1}{2}M\frac{2GM_{J}}{R_{J}} = \frac{GMM_{J}}{R_{J}} Your way works better, and is easier to understand. Thanks :)

After looking it over again, I think maybe I could use an integral. W_{grav} = -Gm_{e}m\int_{r_{1}}^{r_{2}}\frac{dr}{r^2} That is listed in our text. So maybe... W_{grav} = -Gm_{e}m\int_{\infty}^{0}\frac{dr}{r^2} I don't think that's at all right. Just thinking aloud If anyone has any...

Homework Statement A meteor of mass about 3.4 x 10^12 kg is heading straight for Jupiter. When it hits there will be a huge release of energy, visible her on earth. Assuming it has fallen from far away, how much energy will be released when it hits Jupiter? The radius of Jupiter is 7 x 10^7 m...

n_{1}(5)sin64 = (M_{ladder}g)(2.5)cos64 + (M_{weight}g)(5)cos64 n_{1} = 550 F_{tot} = \sqrt{550^2 + 1470^2} = 1570 And 550 < f_{s-max} of 588 :D Thanks for the help!

I_{c} = MR^2 = (3.4)(.53)^2 = 0.95506 I_{d} = \frac{1}{2}MR^2 = \frac{1}{2}(3.4)(.53)^2 = 0.47753 From what you said, I_{d}\omega_{1} = I_{d}\omega_{2} + I_{c}\omega_{2} \omega_{2} = \frac{I_{d}\omega_{1}}{I_{d} + I_{c}} \omega_{2} = \frac{(0.47753)(5.236)}{0.47753 + 0.95506} =...

It just occurred to me that it is asking for the entire force from the ground, not just the frictional force. F_{tot} = \sqrt{n_{2}^2 + f_{s}^2} = \sqrt{1131^2 + 1470^2} = 1852 N Still off =/ *EDIT* using the answers as I guide, I figured out a way to make my calculations match an...