Here is our system. Note the off-center weight acting on the beam.

**Rough estimate**

We expect the three forces to each be less than 10 N, based on some quick trial and error. If the first upwards force is 2 N, the second will be about 8 N (that's 6 N more) and 2 + 8 = 10. Our downward force is 5 times the first, somewhere near the 4 times that what we need, so it's a pretty good guess.

**Solution**

Let the first upward force be *F*, as shown in the diagram.

The statement "the downwards force is four times the first upward force" gives us:

- The downwards force is `4F`, and

The statement "the second of the two upward forces is `6.4\ "N"` more than the first" gives us

- The second upward force is `F+6.4`.

The second sentence in the question tells us that the sum of the 2 upwards forces equals the downwards force. So

`F + (F + 6.4) = 4F`

`2F + 6.4 = 4F`

`6.4 = 2F`

So `F = 3.2`

The second force is `F + 6.4 = 3.2 + 6.4 = 9.6`

The third (downward) force is `4F = 4 xx 3.2 = 12.8`

So the 3 forces are: `3.2` N, `9.6` N and `12.8` N.

This is close to our earlier estimate and all statements in the question are true with these forces.