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LUCKY_DIMON [66]
3 years ago
10

A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 g

ram mass be attached so that the meterstick will be balanced in rotational equilibrium
Physics
1 answer:
alekssr [168]3 years ago
5 0

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                  y cm  

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

              12cm             50 cm              69cm

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                 19 cm                                              

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   

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