Recent content by C. Lee

I am now reading Lagrange's equations part in Taylor's Classical Mechanics text. It says: When a system of interest involves constraint forces, F_cstr, and all the nonconstraint forces are derivable from a potential energy(U), then the Lagrangian for the system L is L = T - U, where U is the...

Hello, I was working on a classical mechanics problem, and I suddenly came up with this question: Let's think about a point in a 2-D plane. This point is represented by (r, θ), where r is the distance between the origin and the point and θ is the angle measured from the x-axis. Now, what is...

Again, |T1| is the magnitude of the tension on Cable 1, not the load on it. Furthermore, each cable does not have to share an equal load with others; those 4 cables could have all different loads on them. For example, if they are set appropriately, Cable 1 can have 500lb for its y-component...

Now you have quadratic equation there, q12+3.56E-10 = 2.3E-4*q1, which you can easily solve.

I don't get how you ended up there. If q1 = 0, q1*q2 cannot have nonzero value.

You do not need to keep indicating they are absolute values since it is clarified in the problem that they are positively charged. Since you have two equations about q1 and q2, you can now solve them easily!

T stands for "the magnitude" of tension. T1, T2, T3, and T4 in bold text stand for tension for each cable. To maintain equilibrium, the sum of all, say, y-components of tensions should balance out the gravitational force on the object.

Yeah but I think a load on a cable is not the magnitude of tension of the cable, but a certain component of it. For example, think about two cables are keeping an object from falling to the ground at the ceiling, one making 30° and another 120° with the x-axis(horizontal to the ground)...

They do not share an equal load. How could they while the components are all different?

I think that is because the definition of R and acceleration are somewhat different in two cases. Acceleration is the rate of change in velocity. But think about R. Is the definition of R the rate of change in V with respect to I? It is the ratio of V to I, not the rate of change. Hope this...

Absolutely right. As you have pointed out from the beginning LHS has extra L/T that RHS does not have, therefore it cannot be dimensionally correct.

You are missing ^2 at the end of the equation: it should be E = (1/2)mv^2.

Right. So the acceleration is constant, and you are given with initial velocity. You can now calculate the velocity at t = 1.0 s.

Alright. So, you have used the gravitational acceleration 9.8m/s^2 to calculate the displacement. What is the definition of acceleration?

So what is your question? Can you give some more details on how you ended up with that answer?