I think the best way to do this is to choose the bottom right corner as the pivot point. Then you get 8*9.8*0.2 = p * 0.4
therefore p = 39.2 for the horizontal force
A sign is supported at its top right corner, point P. The sign is a square with 40 cm on each side and has 8.0 kg mass. What is the magnitude of the horizontal force that P experiences?
2 answers
So I choose the bottom right corner as the pivot point. The radial arm up to point P is the length L of the side of the sign, and the horizontal force is perpendicular to this radial arm. The weight acts straight down, so the radial arm perpendicular to the weight is half the length of the side of the sign. This gives
T(clockwise) = T(counterclockwise)
F(horizontal) * L = w * L/2
Which gives F(horizontal) = mg / 2
My question - aren't we now treating the weight as if it were located at the midpoint of the bottom edge of the sign? I don't mind treating the weight as if it is located at the center of the sign, but located at the center of the bottom edge seems a little odd.
T(clockwise) = T(counterclockwise)
F(horizontal) * L = w * L/2
Which gives F(horizontal) = mg / 2
My question - aren't we now treating the weight as if it were located at the midpoint of the bottom edge of the sign? I don't mind treating the weight as if it is located at the center of the sign, but located at the center of the bottom edge seems a little odd.