Moment of a Force: How to Find an Unknown Force on a Pivoted Metre Stick

[Rumplestiltskin]

Homework Statement

A metre stick is pivoted at 50cm and is in equilibrium.
There is a 2.0N upwards force at the 5.0cm mark, and a 12N downwards force at the 40cm mark. Force F acts at the 75cm mark. Find force F. Assume the upwards direction is positive.

Homework Equations

The Attempt at a Solution

The leftmost upwards force: 2 x 0.45 = 0.9 Nm (as the force would be 45cm from the pivot)
The leftmost downwards force: 12 x 0.1 = 1.2 Nm
Resultant leftmost force: 0.9 - 1.2 = -0.3 Nm

Force F acts 0.15 m from the pivot, therefore: F x 0.15 m = 0.3 Nm. (resultant would be positive in order to counterbalance).
Rearranging for F, 0.3/0.15 = 2 N. The programme marks this as incorrect.

 

[Fightfish]

A metre stick is pivoted at 50cm and is in equilibrium.
Force F acts at the 75cm mark.
Force F acts 0.15 m from the pivot

See anything strange going on there?

 

[Rumplestiltskin]

See anything strange going on there?

Nope. The stick is pivoted at 0.5m. If the force acts at the 0.75m mark, it is acting 0.15m away from the pivot.

EDIT: Ugh, I'm an idiot. Thanks!

 

[Fightfish]

Nope. The stick is pivoted at 0.5m. If the force acts at the 0.75m mark, it is acting 0.15m away from the pivot.
EDIT: Ugh, I'm an idiot. Thanks!

Haha, when I saw your reply earlier without the edit, I was like, "huh"? xD