Solve Problem 13: Magnitude of Reaction Force at Pivot

[Pakbabydoll]

Homework Statement

n my homework we are asked to find the magnitude of the reaction force at the pivot at a instant and earlier in the problem we found the magnitude of the angular speed at one instant, the magnitude of the angular acceleration at the same instant, and the magnitude of the acceleration of its center of mass at the same instant

PROBLEM # 13

Homework Equations

I am lost now... am attaching the picture

The Attempt at a Solution

F= (4.5)(16.43509961) (my numbers)
but the answer is wrong..

than I tried
F= ma
F= (4.5)(7.350000002) (transitional acceleration only)
answer is still wrong :(

my numbers =

angular acceleration= 2.133526851
angular velocity= 2.065684802
Net acceleration= 16.43509961

mass- 4.5
rod- 6.89

What am I doing wrong?

 

[Pakbabydoll]

HEEEEEELPPPPPPPPPPPPPPPPPP please!

 

[Pakbabydoll]

so so for future users this is how you get the answer:

(1/2)(9.8)(m)^2+(1/4)(9.8)(m)^2
than take square root of that


but I still don't know why you get it that way... its too late for me to plug it into the system so I am not sure if its even right but that's how you get it...

 

[Doc Al]

The Attempt at a Solution

F= (4.5)(16.43509961) (my numbers)
but the answer is wrong..

than I tried
F= ma
F= (4.5)(7.350000002) (transitional acceleration only)
answer is still wrong :(

On the right track here. Note that F is the net force. You need to find the force of the pivot on the rod.