# I Force on a rod attached to a pivot.

1. Jan 17, 2017

### albertrichardf

Hello.
Can anyone help me confirm the following from the Feynman tips on physics:

The above diagram shows two rods, with a weight attached (the square). Each rod has a pivot attached, though the pivot on the left is fixed and cannot roll. The weight is 2 kg, and the rod is 0.5 m long. The question is to find the horizontal force on the right pivot when the weight is 0.4 m from the ground. Both rods exert a force such that the net vertical component balances the weight but for the following pretend that the rod with the moving pivot bears all the weight.

My reasoning is as follows: the weight exerts a vertical force on the rod. Part of this force is parallel to the rod and part is perpendicular. The rod transfers any parallel force to the pivot, and rotates under the effect of the perpendicular force. For our purposes, the perpendicular force can be ignored. The vertical component of the parallel force can also be ignored because that is balanced by the ground. The horizontal component of the parallel force is thus what is being looked for.

The parallel component is W cos ø, where W is the weight of the mass (2 * 9.8) and ø is the angle. The horizontal component is W cos ø sin ø. All this comes from trigonometry. Therefore the total horizontal force is about 9.4 N.

However, Feynman reasons that the net force must be parallel to the rod. Because of the weight pushing onto the rod, the rod feels a vertical force, which it exerts on the pivot. The horizontal force on the pivot is also felt by the rod and must be such that the net force is directed parallel to the rod. In other words, the vertical and horizontal forces form right triangle with the net force as the hypotenuse. The horizontal force is then
W tan ø = 14.7

So I got 9.4 but he got 14.7. Can someone explain to me why I am wrong? Also in his explanation, if the rod feels only a vertical force due to the weight, where does the horizontal force on the pivot come from?