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The lever law corresponds to a system exposed to two equal and opposite torques with which the system remains in equilibrium.

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ID:(1457, 0)

Iframe

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ID:(15845, 0)

Description

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Since the torque generated by the gravitational force and the lever arm is

on each side of the balance, it must cancel out in the case of equilibrium to achieve balance:

If we assume that on one side we have the force 1 ($F_1$) and the force - axis distance (arm) 1 ($d_1$), and on the other side the force 2 ($F_2$) and the force - axis distance (arm) 2 ($d_2$), we can establish the well-known lever law as follows:

ID:(15847, 0)

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Parameters

$d_1$

d_1

Force - axis distance (arm) 1

m

$d_2$

d_2

Force - axis distance (arm) 2

m

$F_1$

F_1

Force 1

N

$F_2$

F_2

Force 2

N

$g$

g

Gravitational Acceleration

m/s^2

$m_1$

m_1

Mass 1

kg

$m_2$

m_2

Mass 2

kg

$T_1$

T_1

Torque 1

N m

$T_2$

T_2

Torque 2

N m

Variables

Calculations

• Alternative method
• Traditional method
• Strategy
• 2D
• 3D

Equation

Number

First, select the equation: to , then, select the variable: to

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Calculations

Symbol

Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable Given Calculate Target : Equation To be used

Equations

ID:(15846, 0)

Equation

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If a bar mounted on a point acting as a pivot is subjected to the force 1 ($F_1$) at the force - axis distance (arm) 1 ($d_1$) from the pivot, generating a torque $T_1$, and to the force 2 ($F_2$) at the force - axis distance (arm) 2 ($d_2$) from the pivot, generating a torque $T_2$, it will be in equilibrium if both torques are equal. Therefore, the equilibrium corresponds to the so-called law of the lever, expressed as:

$d_1 F_1 = d_2 F_2$

Conditions

Variables

$d_1$

Force - axis distance (arm) 1

$m$

6138

$d_2$

Force - axis distance (arm) 2

$m$

6139

$F_1$

Force 1

$N$

6140

$F_2$

Force 2

$N$

6141

Relation

Development

In the case of a balance, a gravitational force acts on each arm, generating a torque

$T = r F$

If the lengths of the arms are $d_i$ and the forces are $F_i$ with $i=1,2$, the equilibrium condition requires that the sum of the torques be zero:

$\displaystyle\sum_i \vec{T}_i=0$

Therefore, considering that the sign of each torque depends on the direction in which it induces rotation,

$d_1F_1-d_2F_2=0$

which results in

$d_1 F_1 = d_2 F_2$

.

ID:(3250, 0)

Equation

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Since the relationship between angular momentum and torque is

its temporal derivative leads us to the torque relationship

$T_1 = d_1 F_1$

$T = r F$

Conditions

Variables

$F$

$F_1$

Force 1

$N$

6140

$r$

$d_1$

Force - axis distance (arm) 1

$m$

6138

$T$

$T_1$

Torque 1

$N m$

10410

Relation

Development

Si se deriva en el tiempo la relación para el momento angular

$L = r p$

para el caso de que el radio sea constante

$T=\displaystyle\frac{dL}{dt}=r\displaystyle\frac{dp}{dt}=rF$

por lo que

$T = r F$

The body's rotation occurs around an axis in the direction of the torque, which passes through the center of mass.

ID:(4431, 1)

Equation

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Since the relationship between angular momentum and torque is

its temporal derivative leads us to the torque relationship

$T_2 = d_2 F_2$

$T = r F$

Conditions

Variables

$F$

$F_2$

Force 2

$N$

6141

$r$

$d_2$

Force - axis distance (arm) 2

$m$

6139

$T$

$T_2$

Torque 2

$N m$

10411

Relation

Development

Si se deriva en el tiempo la relación para el momento angular

$L = r p$

para el caso de que el radio sea constante

$T=\displaystyle\frac{dL}{dt}=r\displaystyle\frac{dp}{dt}=rF$

por lo que

$T = r F$

The body's rotation occurs around an axis in the direction of the torque, which passes through the center of mass.

ID:(4431, 2)

Equation

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The gravitational Force ($F_g$) is based on the gravitational mass ($m_g$) of the object and on a constant reflecting the intensity of gravity at the planet's surface. The latter is identified by the gravitational Acceleration ($g$), which is equal to $9.8 m/s^2$.

Consequently, it is concluded that:

$F_1 = m_1 g$

$F_g = m_g g$

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

5310

$F_g$

$F_1$

Force 1

$N$

6140

$m_g$

$m_1$

Mass 1

$kg$

10412

Relation

ID:(3241, 1)

Equation

>Top, >Model

The gravitational Force ($F_g$) is based on the gravitational mass ($m_g$) of the object and on a constant reflecting the intensity of gravity at the planet's surface. The latter is identified by the gravitational Acceleration ($g$), which is equal to $9.8 m/s^2$.

Consequently, it is concluded that:

$F_2 = m_2 g$

$F_g = m_g g$

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

5310

$F_g$

$F_2$

Force 2

$N$

6141

$m_g$

$m_2$

Mass 2

$kg$

10413

Relation

ID:(3241, 2)