Recent content by danielu13

How does surface area affect pressure, since pressure is defined by \frac{Force}{Area}, and specifically how does this work with the Ideal Gas Law (PV = nRT)? I would think that surface area and pressure have an inverse relationship, as to pressure and volume. But what if you had gas-filled...

Well, based on what you said, I was thinking that the tension is based on the even distribution of the masses, which is why I'm thinking the average of the masses. But I have some feeling that it should be g(m2-m1) since the pulley makes the gravitational forces counteract each other. Am I at...

These pulley problems have always confused me for some reason. If there were two unequal masses on a pulley would the tension be T = g\frac{m_1+m_2}{2} ?

I was familiar with the use of complex numbers in solving the ODE, but not with relating to the rest of the concepts of SHO. That link gave me a bit more understanding. Thanks!

Yes, the tension in the string is constant. However, there are two components to the tension in the string; one is caused by the force of gravity acting on mass 2 and the other is the force of friction acting on mass 1. I may have issues where I used a positive instead of negative. Tension is...

Well what you have is roughly 15 foot-pounds of torque, which I believe most electrical motors should be easily capable of.

As the others have said, it really depends on your professor and college. I'm taking those three classes in addition to a psychology class. I'm doing fairly well except for physics, but no one I have talked to is doing well in that class. I would say if you can to Calc 2, the others shouldn't be...

For the Tension, you calculate the amount of tension in each section of string and add them together. The tension in T1 is calculated by the force of friction with is F_{friction} = F_N\mu_k For this, you use the coefficient of kinetic friction, as I am assuming that the force is being...

Okay, that's what I was thinking but I saw that both A and \delta were related to the constants, and thought there may be some way to relate the two.

In the problems that I have dealt with, I've been given the A value and had to calculate \delta. I was trying to see if there is a simple way to relate it without solving every time, but the more I think about it, the less likely it seems.

So the cos(D) and -sin(D) represent the phase shift, and that's where the delta term comes from in the generalized solution?

The way that I would do it would be to rewrite the equation as \mu^2 '' = -ae^\mu Then you solve for a general solution which is \mu(x) = 1 + x Then you go on to get your general solution from the boundaries and such. Another method, which would probably be easier and is what I think...

Those are constants in the solution to the differential equation that are determined by the initial conditions.

I'm trying to work out the differential equation for simple harmonic motion without damping, x''+\frac{k}{m}x = 0 I can solve it to x = c_1cos(\sqrt{\frac{k}{m}}) + c_2sin(\sqrt{\frac{k}{m}}) But the generalized solution is x = Acos(\omega*t + \delta) where A = \sqrt{c_1^2 + c_2^2}...

So, simple harmonic motion without damping is described generally by x(t) = Acos(\omega*t +\delta) Which is derived from the differential equation x''+\frac{k}{m}x = 0 We know that A = \sqrt{c_1^2+c_2^2} and tan\delta = \frac{c_1}{c_2} With the differential equation, dealing...