a uniform rod of 30cm is pivoted at its center. a 40N weight is hung 5cm from left. from where 50N weight be hung to maintain equilibrium?

The 50N weight be hung  at 23 cm to maintain equilibrium

Explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

     05cm15cmP30cm

              ↓             10cm      Δ       xcm    ↓

             40N                                             50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung  at 23 cm to maintain equilibrium

the independent variable, or the one that is intentionally manipulated is the speed.

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.