A meter stick on earth made from a uniform-density material rests on a fulcrum positioned precisely at its middle. If a box of mass m = 3 kg

Question

A meter stick on earth made from a uniform-density material rests on a fulcrum positioned precisely at its middle. If a box of mass m = 3 kg is suspended from the right end of the meter stick, the system would not be in equilibrium. Where and how could a force be applied to keep the system in equilibrium? (For ease of calculation, consider g = 10 N/kg.)

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Ben Gia 4 years 2021-07-26T16:15:37+00:00 1 Answers 40 views 0

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  1. Eirian
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    2021-07-26T16:17:28+00:00
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    Answer:

    A force of 75 N placed at 0.7 m on the meter stick.

    Explanation:

    The weight of the box is equal to:

    W=mg=3*10=30N

    The net torque is equal to zero and is equal to:

    F(x-0.5)-(30*0.5)=0\\F(x-0.5)=15

    For a force with a value of 75 N that is placed at 0.70 m on the meter stick, it would produce a torque of 15 N m

    If you replace that values in the equation:

    75(0.7-0.5)=15\\15=15

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