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.