Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 weeks ago

A 56 kg uniform board 2.36 m long is supported by a pivot 79 cm from the left end and by a scale at the right end?

A 56 kg uniform board 2.36 m long is supported by a pivot 79 cm from the left end and by a scale at the right end (see image).

1. Where should the 43 kg child sit if the scale is to read 100 N? in cm FROM THE LEFT END OF THE BOARD.

2. Where should the 43 kg child sit if the scale is to read 300 N? in cm FROM THE LEFT END OF THE BOARD.

Attachment image

2 Answers

Relevance
  • 4 weeks ago

    This also is the same question that I answered both two days ago and also today https://answers.yahoo.com/question/index?qid=20191...

    and https://answers.yahoo.com/question/index?qid=20191...

  • 4 weeks ago

    that means there is a force of 100 N downwards on the right end.

    CW torques:

    1. board itself, on the right is 2.36 – 0.79 = 1.57 m and the center of that section is half way from the pivot. Half is 0.785 m from the pivot. Weight of that section is (1.57/2.36)56 = 37.25 kg. so torque is 37.25 kg x 0.785 m = 29.24 kg-m

    2. force from scale = 100 N x 1.57 m = 157 Nmconverting to kg-m that is 157/9.8 = 16.02 kg-m(technically, I should convert all the torques to Nm but it is less work using kg-m)

    CCW torques

    1. board itself. On the left is 2.36-1.57 = 0.79 m and the center of that section is half way from the pivot. Half is 0.395 m from the pivot. Weight of that section is (0.79/2.36)56 = 18.75 kg. so torque is 18.75 kg x 0.79 m = 14.81 kg-m

    2. boy, 43 kg x D m = 43D

    set torques equal

    29.24 + 16.02 = 14.81 + 43D

    43D = 30.45

    D = 0.71 m

    that is from pivot.

    from left end it is 0.79 – 0.71 = 0.08 m

    (2) is the same, just change a number....

Still have questions? Get your answers by asking now.