# Force by table leg

1. Homework Statement
https://courses.edx.org/static/content-mit-mrev~2013_Summer/problems/MIT/rayyan/check_points/Pictures/BK85.png [Broken]

A horizontal uniform bar of mass 220 kg and length L = 3.0 m is placed on two supports, labeled A and B, located as shown in the diagram. A block of mass 30 kg is placed on the right end of the bar.

Find the normal forces F_A and F_B exerted on the bar by the supports.

2. Homework Equations
I'm sure we'll use torque, T = rFsinθ. We'll probably use center of mass, Σmx / M. Because the bar is uniform, though, we know the center of mass will be the mid-point of the bar. Torque will be determined by the gravity acting on the center of mass, which has equation F = mg. Since the object is static the sum of all torques will be 0.

3. The Attempt at a Solution
I treat the point B as if it were a pivot point, in which case the distance to the center of mass is 0.5m to the left. Then the torque applied is T1 = 0.5 * 220g = 110g. The block has distance 1m to the right of the pivot and so exerts a torque T2 = -1*30g in the opposite direction. The normal force due to support A will then satisfy 110g-30g+N=0 and this implies N = -80g.

However, this answer turns out to be wrong. Any idea what my mistake is?

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#### haruspex

Homework Helper
Gold Member
2018 Award
The normal force due to support A will then satisfy 110g-30g+N=0 and this implies N = -80g.
Are you defining N to be the force at A or the torque it exerts about B?

Oh wow, can't believe how simple my mistake was. Thanks for the hint, haruspex!

"Force by table leg"

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