A meterstick of negligible mass is placed on a fulcrum at the 0.4 m mark, with a 1 kg mass hung at the zero mark and a 0.5 kg mass hung at the 1.0 m mark. The meterstick is held horizontal and released. Immediately after release, the magnitude of the net torque on the meterstick about the fulcrum is most nearly: a. 1 Nm
b. 2 Nm.
c. 2.5 Nm.
d. 7 Nm
e. 7.5 Nm

The net torque is 0.98 Nm.

Explanation:

The torque is given by

Torque = force x perpendicular distance

The clock wise torque is taken as negative while the counter clock wise torque is taken as positive.

Take the torques about the fulcrum.

Torque =  1 x 9.8 x 0.4 - 0.5 x 9.8 x 0.6

Torque = 3.92 - 2.94 = 0.98 Nm

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.