Posted by Latane on .
A uniform horizontal rod of mass 2.4 kg and
length 0.86 m is free to pivot about one end
as shown. The moment of inertia of the rod
about an axis perpendicular to the rod and
through the center of mass is given by I =
(m*l^2)/12
If a 6.8 N force at an angle of 69 to the hor
izontal acts on the rod as shown, what is the
magnitude of the resulting angular acceleration about the pivot point? The acceleration
of gravity is 9.8 m/s2 .
Answer in units of rad/s2

physics 
bobpursley,
I don't understand the force placement.
Use Torque=(momentinertia)angacc
to get the moment of inertia, use the Parallel axis law.
To get torque, force*distance*sinAngle