Recent content by smashd

Meh... The density of copper was something that's just supposed to be looked up. I thought I was supposed to solve for it another way. It's 8.96 g/cc. Never mind, problem solved!

Question: A copper wire and steel wire with identical diameters are placed under identical tensions. The frequency of the third resonant mode for the copper wire is found to be the same as the frequency of the fourth resonant mode for the steel wire. If the length of the copper wire is 3.44 m...

Yeah, there's a lot of squaring and factoring going on. But I tend to over think on these algebra problems, so I might have taken the long way around. Either way, I've attached a picture of my work to this post. I hope it helps... M is ## m_{sample} ## and m is ## m_{rope} ##.

I solved it. I thought I'd share my solution for future reference. Thanks again for the help! v(y) = \sqrt{\frac{T(y)}{μ}} \frac{dy}{dt}\ = v(y) \sum F_{y} = T(y) = M(y) ~ g -------------------- T(y) = M(y) ~ g M(y) = m_{sample} + \frac{m_{rope}}{d} ~ y T(y) = \left(m_{sample} +...

Changed it to y so that's it's a little more intuitive. Mass of the system: M(y) = m_{sample} + \frac{m_{rope}}{d} ~ y Tension on the system: T(y) = M(y) ~ g = (m_{sample} + \frac{m_{rope}}{d} ~ y) ~ g Wave speed: v(y) = \sqrt{\frac{T(y)}{μ}} = \sqrt{\frac{(m_{sample} +...

Thanks for the replies. Hmm, in what way? I already know that ## v = \sqrt{\frac{T}{μ}} ##. Well I know that the tension at the top is T = (m_{samples} + m_{rope})~g , so that v = \sqrt{\frac{(m_{samples} + m_{rope})~g}{\frac{m_{rope}}{d}}} . The last equation I just wrote isn't a...

Homework Statement A geologist is at the bottom of a mine shaft next to a box suspended by a vertical rope. The geologist sends a signal to his colleague at the top by initiating a wave pulse at the bottom of the rope that travels to the top of the rope. The mass of the box is 20.0 kg and the...

Homework Statement A diving board of length L is supported at a point a distance x from the end, and a diver weighing w1 stands at the free end (Figure 1) . The diving board is of uniform cross section and weighs w2. (Figure 1) Find the force at the support point. Find the force at the end...

Thanks for the input, WillemBouwer. So \sum F_{3y} should be: \sum F_{3y} = m_{3}g - T_{1} = m_{3}a Then a would become after combining the forces on the system: \frac{m_{3}g - m_{2}g\sin\theta - m_{1}g\sin\theta}{m_{1} + m_{2} + m_{3}}Which is still 2.1 m/s^2, but this is the proper...

Homework Statement A system comprising blocks, a light frictionless pulley, a frictionless incline, and connecting ropes is shown. The 9 kg block accelerates downward when the system is released from rest. The acceleration of the system is closest to: A.) 1.9 m/s^2 B.) 2.1 m/s^2 C.) 1.7 m/s^2...