Recent content by Jennings

First attachment is solving for the angle which I found to be 86.4 degrees. (It really doesn't look that way in the picture I hope I didn't do it wrong). Second attachment is the sum of forces in direction x which is also the magnitude of the centripetal force

Chet Something else needs to be known. The sum of forces in the y direction must be 0 so that we can have an equation. (Fs means force of spring) ΣFy = Fscosθ - mg ΣFx =Fssinθ If the sum of forces in the y direction = 0 then we have our equation. Otherwise i don't kneo

Wait hold on I think I got something

Sounds good. Only there's one problem. I don't know the direction of the spring force, only its magnitude. What I know : Magnitude and direction of the weight Magnitude of the spring force Don't know: Direction of spring force Magnitude and direction of resultant centripetal force. What...

I thought that the spring force is the net force? That's how I was able to find the net force. Please clarify

Here is my attempt at solving for the centripetal force. I don't think its correct but it's all I know how to do

Hello Chestermiller. In order to solve for angle theta Atleast one more component of the force must be known. Weight is straight down and I'm guessing the centripetal force must make up rest of the net force. Isn't centripetal force is purely horizontal?

I don't know how to attach the file of my free body diagram

1. Homework Statement (Attached is an image depicting the situation) A ball of mass m = 1.50kg is hung from a spring attached to a shaft. The length of the spring with the mass hanging from it is Lo = 50.0cm. The shaft then starts to rotate such that the spring stretches to a length L =...

Wow that's crazy. This problems not suppose ti be that hard

Yay! Thank you! Your a great helper this is the 2nd time in less than a week. Do you get points for this?

Kinetic friction between M2 and surface is (.25)(147N) = 36.8N Kinetic friction between M1/M2 = (.25)(49.1N) = 12.3N The sum of forces acting on M2 F - Fk1 - Fk2 = 50N -36.8N - 12.3N = 0.9N 0.9N = M2(a) 0.9N = (10kg)a a = .09m/s^2 Is it correct? Tension = Fk(mass1/2) because mass...

Assuming the friction is kinetic .. do you have any hints you can give to me to set me up on the right path. I am assuming that value for kinetic friction will be less with M1 since M1 has a smaller normal, but M1 itself does not move, so its kind of confusing on what exactly to do