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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.)

Respuesta :

Answer:

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

Explanation:

The weight of the box is equal to:

[tex]W=mg=3*10=30N[/tex]

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

[tex]F(x-0.5)-(30*0.5)=0\\F(x-0.5)=15[/tex]

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:

[tex]75(0.7-0.5)=15\\15=15[/tex]

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