A hungry bear weighing 720 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 5.40 m long; the basket weighs 80.0 N.
a) When the bear is at x = 1.00 m, find the tension in the wire and the components of the force exerted by the wall on the left end of the beam.
N ( Fx)
N ( Fy)
The Attempt at a Solution
Okay well I sort of understood this question and created a balanced torque equation. After doing this I got that:
Tension = 345.66
Fx = 172.83
But I need help with finiding Fy Please.
I only have like one more chance to answer the question on my homework so I want to make sure I can get the right answer.
I know I have to write out some equation like maybe:
Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)
Is this correct? And if its not, am I on the right track?