I think you made a mistake or I am not understanding this(Around 8:20). For
torque, shouldn’t r = .5m for the cat and the board since the axis of
rotation is right in the middle? So the answer would the mass of the cat is
equal to the mass of the board.

Why is the mg positive for the force acting on the board if the mg for the
cat is negative? You said that the mass times acceleration of gravity
rotates clockwise for both the cat and the board so they have negative
torques. Does a negative torque indicate a negative value when you
substitute the forces into the net force equation and set it equal to zero?
Or is the mg for the board positive because the board would go up, from the
reference point of the ground, if the mg for the cat is greater, and that
side would go down?

the answer i got is mcat=mboard 0.33333 so mboard is the kg?

Very Helpful!

would you subtract Mbg because it turns clockwise direction?

Nevermind, I found out that board was 4 meters.

is it 2.33333333333333333 kg?

what was the answer?

Why isnt the normal force acting from the board to the cat not included?

why is there no normal force from the other sawhorse

I think you made a mistake or I am not understanding this(Around 8:20). For

torque, shouldn’t r = .5m for the cat and the board since the axis of

rotation is right in the middle? So the answer would the mass of the cat is

equal to the mass of the board.

Why is the mg positive for the force acting on the board if the mg for the

cat is negative? You said that the mass times acceleration of gravity

rotates clockwise for both the cat and the board so they have negative

torques. Does a negative torque indicate a negative value when you

substitute the forces into the net force equation and set it equal to zero?

Or is the mg for the board positive because the board would go up, from the

reference point of the ground, if the mg for the cat is greater, and that

side would go down?

nopr, its nawt 🙂

did it bug anyone else that the entire video, the cat’s back was not

connected?