Recent content by snormanlol

Hi bVU, Thanks for the tip. arad= v^2/L. So that a = 2*g*(1-cos(theta)). T - mg = 2*m*g*(1-cos(theta)) => T =mg(3-2*cos(theta))

For part a I used conservation of energy. -m*g*cos(θ)*L+1/2*m*0^2=-m*g*L +1/2*m*v^2 => v = sqrt(2*g*L(1-cos(θ )). b) For b I was think that T = mg in the equilibrium point but that doesn't invole θ in the answer. So that's why I tought that T*cos(θ ) = mg. So that the tension is mg/cos(θ). But...

Ah yes it misleaded me thank you guys and yeah I meant arccos but I made a mistake. But thank you for the help guys.

Well we can take Arctan(1/1.5) but I don't see why we can say that the perimeter is 5. Isn't the loop 5 m? And the length of the sides are unknown? That's why I tried using torque but I got the equation 2*T*sin(theta)=147.25 back.

I know that the answer has to be 98.6N. So I know that Fy=0 so that 2*T*sin(theta) = 147.25. Then I was think to take the torque of the left wheel but I can't find the lever arm of the tension force. I also know that u can solve the question by saying that the 2 sides of unknown length are 1.5 m...