Question

На горизонтальный рычаг подвешены два груза. Первый груз массой 10 кг создаёт на рычаге момент силы 200 Н*м. На каком расстоянии от точки опоры рычага нужно подвесить второй груз массой 2 кг, чтобы рычаг находился в равновесии?

Answer 1

M1=10 кг    m2=2 кг   M1=200 Н*м    L2=?
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M1=M2
M2=m2*g*L2
L2=M1/(m2*g)=200/(2*10)=10 м
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Answer 2

Problem Statement

We are given a horizontal lever with two hanging masses. The first mass of 10 kg creates a torque of 200 N*m on the lever. We need to find the distance from the lever's fulcrum where the second mass of 2 kg should be hung so that the lever is in equilibrium.

Analysis

Let's think step-by-step: * The first mass (m1) of 10 kg is creating a torque (T1) of 200 N*m on the lever * Torque is defined as the product of force and perpendicular distance from the fulcrum: T = F × d. * The force due to the first mass is its weight: F1 = m1 × g = 10 × 10 = 100 N. * Setting the torques equal: T1 = T2 * So: F1 × d1 = F2 × d2 * F1 = 100 N (from above), F2 = m2 × g = 2 × 10 = 20 N * T1 = 200 N*m = 100 N × d1 (from definition of torque) * T2 = F2 × d2 = 20 N × d2 (definition of torque for second mass) * Equating the torques: 100 × d1 = 20 × d2 * Solving for d2: d2 = 100 × d1 / 20 = 5 m

Therefore, the distance from the fulcrum where the second mass of 2 kg should be hung is 5 m.