The meterstick shown to the right is 100 cm long. It is free to pivot around its

Question

The meterstick shown to the right is 100 cm long. It is free to pivot around its center of gravity (CG), which is at the 50-cm mark. There is a 25.0-N block hanging from the 80-cm mark. Decide where each of the other blocks should be placed, one at a time, to balance out the 25.0-N block. At what mark on the meter stick would you place a 19.0-N block to balance the 25.0-N block?

At what mark on the meter stick would you place a 41.0-N block to balance the 25.0-N block?

The meterstick shown to the right is 100 cm long. It is free to pivot around its center of gravity (CG), which is at the 50-cm mark. There is a 25.0-N block hanging from the 80-cm mark. Decide where each of the other blocks should be placed, one at a time, to balance out the 25.0-N block 20 30 40 50 60 70 80 90 25.0 N 19.0 N At what mark on the meter stick would you place a 19.0-N block to 41.0 N alance the 25.0-N block? Number cm At what mark on the meter stick would you place a 41.0-N block to balance the 25.O-N block? Number cm

Explanation / Answer

Given that the length of the meter stick is 100 cm.

Now, right-side torque = F * L = 25 * (80- 50) = 750 N/cm

(a) To balance with 19 N block -

Required distance at which it will be placed = 50 - (750/19) = 10.53 cm. mark.

(b) To balance with 41 N block -

Required distance at which it will be placed = 50 - (750/41) = 31.70 cm. mark.

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