Recent content by HoodedFreak

I believe that should give you a = g(1 - 3μ)/4 Then you need to ensure that a is positive, g and 1/4 are positive constants so you have the restriction that: 1-3μ ≥ 0

I was rushing everything and made a lot of mistakes. I corrected those mistakes since then and got the correct answer. Thanks for the help!

The change in length should be Δl Using the two equations I had: T - 12g = 12*16π / (0.7 + Δl) and Y * Strain = Stress ⇒ 6.9*10^9 * Δl/0.7 = T/0.014 Solving for T from the first equation to get T = 12*16π / (0.7 + Δl) + 12g Plugging it into the second, to get 12*16π / (0.7 + Δl) + 12g =...

Also, to harupex's point, should I ignore the elongation and just treat the radius of the circle as R = lo = 0.7m since it should be small in comparison to the length.

Great thanks for the help!

Right, yeah, so that would be 69*10^9 then. Okay, so if I plug in 69*10^9 * Δl/lo = T/0.014, then T = 69*10^9*0.014 * Δl / 0.7 = 1380,000,000*Δl SO 1,380,000,000*Δl - 12g = 12*16π / (0.7 + Δl) (1,380,000,000*Δl - 12g) * (0.7 + Δl) = 603 And I solve this quadratic? It seems like I'm doing...

Right, okay, so the center of gravity in between the two balls should be at a horizontal distance of 1.5 from the corner, which is 0.5 from point B, so the torque due to grabity would be 0.5*0.15g, so in the end we have 0.075g + Fa = (√3 + 1)Fa, which gives me 0.075g = √3Fa, and in the end, We...

Young's modulus for aluminum, according to google is 69 N/m^2

Homework Statement A 12.0-kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.70 m, is whirled in a vertical circle with a constant angular speed of 120 rev>min. The cross-sectional area of the wire is 0.014 cm2. Calculate the elongation of the wire when the mass is...

I see, that makes perfect sense thanks for the explanation. Okay, so now I form the triangle with horizontal 1 and hypotenuse 2, solving for the height gives me √(22 - 12 = √3. So a total vertical distance of √3 + 1. So the torque on the system from point C woould be Fc * (√3 + 1). So i get...

Why did you draw a straight line between the two centers, isn't it possible to have the two centers not lie on a line through both their radiuses, I'd draw what I mean, but paint won't allow me to, something along these lines

How do you know what the center to center distance is? Clearly the horizontal distance of the centers would be 1, but how do you know the distance from one center to another?

I see, okay. But why is it valid to treat them both as one system together since they are not attached to each other? Okay, so if I treat them as a system together, then the lever arm at point C would simply be the vertical distance to point C, but we aren't given how high above the bottom of...

My last message is complete I just forgot to put the period. Which two torques are zero about point B? And why are we looking for the lever arm of point C when point C is not in contact with the bottom left marble.

Okay, well if I look at the torque of the first marble about point A, we have the torque from the force of the wall at A being zero, then we have the torque of the Force B being 0.15*g*1, and the torque of the second marble acting on the first, which I don't know how to find