Question

Невесомый рычаг, имеющий опору в точке, делящей рычаг в отношении 1:5, находится в равновесии при подвешивании к короткому концу груза массой 5 кг и некоторого груза к другому концу. Определить вес груза, прикрепленного к длинному концу рычага.

Answer 1

F1d1=F2d2
d1=x м, d2=5x м
F1=mg=5*10=50 H
F2=F1d1/d2=50*x/5x=10 H

Answer 2

$M= F_{m} l \\ M_{1} = M_{2} \\ F_{m} =mg \\ m_{1} g l_{1} = m_{2} g l_{2} \\ l_{2} =5 l_{1} => \\ m_{2} g= \frac{ m_{1} g l_{1} }{5 l_{1} } \\ m_{2} g=10H$

Answer 3

Solution:

To solve this problem, we can use the principle of moments, which states that for an object to be in equilibrium, the sum of the anticlockwise moments about any point must be equal to the sum of the clockwise moments about the same point.

Let's denote the weight of the load attached to the short end of the lever as W1 and the weight of the load attached to the long end as W2.

Given: - The lever is divided in the ratio 1:5. - The weight of the load attached to the short end, W1, is 5 kg.

We need to find the weight of the load attached to the long end, W2.

Using the principle of moments, we can set up the equation:

W1 × d1 = W2 × d2

Where: - d1 is the distance from the pivot to the point where the load of 5 kg is attached. - d2 is the distance from the pivot to the point where the unknown load is attached.

Since the lever is divided in the ratio 1:5, we can express d1 and d2 in terms of the total length of the lever, L: - d1 = L / 6 - d2 = 5L / 6

Now we can solve for W2:

W1 × (L / 6) = W2 × (5L / 6)

W2 = (W1 × (L / 6)) / (5L / 6)

W2 = W1 / 5

Answer:

The weight of the load attached to the long end of the lever is 1 kg.