How Do You Calculate Angular Acceleration in a Pivoting Rod Problem?

[luckysword12]

Homework Statement

A uniform horizontal rod of mass 2.8 kg and
length 2.1 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 4.8 N force at an angle of 63◦
to the horizontal 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/s^2
.
Answer in units of rad/s^2

Homework Equations

Sum of the torques = I * a
Torque = R * F * sin(theta)

The Attempt at a Solution

Here is what I have so far:

(sum of the torques) = (Torque of force F)
I * a = R * F * sin(63)
a = (R * F * sin(63)) / I

I = (M * L^2)/3 (It is rotating at its endpoint not at its center of mass)

a = (3 *(R * F * sin(63))) / (M * L^2)
a = (3 *(2.1 * 4.8 * sin(63))) / (2.8 * 2.1^2)
a = 2.1821 rad/s^2 <-----But this answer is incorrect. Where did I go wrong?

 

[luckysword12]

Sorry if you came to this but I just figured it out. I forgot the torque due to gravity