Calculating Force of Beam on Support at A

[huskydc]

A man of mass mm = 95 kg decides to paint his house. To do this, he builds a platform using a uniform beam with a mass of mb = 100 kg and a length of L = 7 meters. The beam is supported by two sawhorses, as shown in the diagram above.

If the man stands over the support at point B, calculate the force exerted by the beam on the support at A.

--at first, i misread the question and thought it's asking for the force exerted by A on the beam.
and I did the following:

F(A) + F(B) - F(b) - F(m) = 0

and have B as the pivot,

thus with the torque of A = -3 F(A)
and torque of beam (b) = .5 F(b)

, but now it's asking of the F exerted by beam on support at A? I'm confused here...its one whole beam, I'm assuming the force excerted by the beam will occur at the center of mass...help

 

[zwtipp05]

The force of the beam on A is the same as the force of A on the beam. (For every action there is an equal and opposite reaction)

 

[huskydc]

yea, i initially thought it's asking for F(A), and i tried solving for it, didnt work out, unless i did something wrong in my original calculations...

 

[Fermat]

I can't understand why you've been asked the same question twice, but disregrding that, there's a mistake in caculating the torque of the beam.
The beam is a UDL, not a point load acting at its mid-point.
You have to consider the torque from the beam as being up of two parts. one to the left of point B and one to the right.

 

[zwtipp05]

There are two forces of the beam. Since the beam is uniform we can say that one meter has a mass of 100/7 kg. Since two meters are located on the other side of B from the other 5 meters of the beam, you must take that into account as well.

We'll designate the 5 meters to the left of B as b1. The two meters to the right of B shall be called b2.

The problem can be set up as total torque being zero (since it isn't moving). There for the expression for the torque caused by each point is: t(b1)-(t(A)+t(b2))=0. Remember that the torque resulting from the weight of a section of beam essentially acts midway between the end of the beam and the pivot point.