Предмет: Физика, автор: islamirpiisov

Рычаг, длина которого равна 90 см (см. рисунок). находится в равновесии. На каком расстоянии от оси вращения (точка О) подвешен второй груз? Мас- сы грузов равны соответственно 8 кг и 12 кг. Мас- сой рычага можно пренебречь. пожалуйста скажите!!
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Ответы

Автор ответа: shahbazovelman97

1

Объяснение:

Чтобы рычаг находился в равновесии, моменты сил, действующих на него, должны быть равны. Момент силы определяется как произведение силы на расстояние от оси вращения до линии действия этой силы

M1 = 8 кг * 9.8 м/c² * x

M2 = 12 кг * 9.8 м/c² * (90 см - x)

Так как рычаг находится в равновесии, M1 должен быть равен M2:

8 кг * 9.8 м/c² * x = 12 кг * 9.8 м/c² * (90 см - x)

Решая это уравнение относительно x, получаем:

x = 54 см

Таким образом, второй груз должен быть подвешен на расстоянии 54 см от оси вращения.

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