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 centripital 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
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 centripital force.
What...
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 centripital force must make up rest of the net force. Isn't centripetal force is purely horizontal?
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 =...
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