Simple machines. Efficiency

The lever, the arm and the moment of a force, pulleys, the inclined plane, the "golden rule" of mechanics, useful and spent work, efficiency.

Terms

Studied 0 of 10

Formulas

Rule of the lever's equilibrium

F_1 l_1 = F_2 l_2

Quantities
Symbol	Name	Unit
$F_1$	the first force acting on the lever	N
$l_1$	arm of the first force	m
$F_2$	the second force acting on the lever	N
$l_2$	arm of the second force	m

Solved for another quantity:

F_2 = \dfrac{F_1 l_1}{l_2}

\dfrac{F_1}{F_2} = \dfrac{l_2}{l_1}

l_2 = \dfrac{F_1 l_1}{F_2}

Note. The lever is in equilibrium if the forces are inversely proportional to the arms. For example, a force of $40\ \text{N}$ acts on an arm of $0.5\ \text{m}$ ; it can be balanced on an arm of $2\ \text{m}$ by a force $F_2 = 40\cdot 0.5 / 2 = 10\ \text{N}$ — a gain in force of $4$ times.

Moment of a force

M = F\cdot l

Quantities
Symbol	Name	Unit
$M$	moment of the force	N·m
$F$	the force that turns the body	N
$l$	arm of the force	m

Solved for another quantity:

F = \dfrac{M}{l}

l = \dfrac{M}{F}

Note. The lever is in equilibrium if the moment of the force turning it clockwise equals the moment of the force turning it counterclockwise: $M_1 = M_2$ . For example, $20\ \text{N}$ on an arm of $0.5\ \text{m}$ gives $M = 20\cdot 0.5 = 10\ \text{N·m}$ .

Efficiency of a machine

\eta = \dfrac{A_\text{полезн}}{A_\text{затр}}\cdot 100\%

Quantities
Symbol	Name	Unit
$\eta$	efficiency (in per cent)	%
$A_\text{полезн}$	useful work	J
$A_\text{затр}$	spent (total) work	J

Solved for another quantity:

A_\text{полезн} = \dfrac{\eta\, A_\text{затр}}{100\%}

A_\text{затр} = \dfrac{A_\text{полезн}}{\eta}\cdot 100\%

Note. The efficiency of a real machine is always less than $100\%$ . For example, a $10\ \text{kg}$ load is lifted by $2\ \text{m}$ along an inclined plane $4\ \text{m}$ long with a force of $80\ \text{N}$ ( $g = 10$ ): $A_\text{useful} = mgh = 200\ \text{J}$ , $A_\text{spent} = Fl = 320\ \text{J}$ , $\eta = 200/320\cdot 100\% = 62.5\%$ .

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A lever with two forces

The forces are perpendicular to the lever, so an arm is the distance from the support to the point where the force is applied. The weight of the lever itself is ignored.

The lever is in equilibrium: $M_1 = M_2$ .

Graphs

M(F_2)

Blue line — $M_2 = F_2 l_2$ , yellow horizontal — $M_1$

M_1

30 N·m

M_2

30 N·m

F_2\ (\text{to balance})

10 N

Formulas with your numbers

M_1 = F_1 l_1 = 20\cdot 1.5 = 30\ \text{N·m}

M_2 = F_2 l_2 = 10\cdot 3 = 30\ \text{N·m}

F_2 = \dfrac{F_1 l_1}{l_2} = \dfrac{20\cdot 1.5}{3} = 10\ \text{N}

How to read the graphs

Lever: the graph of the moment $M_2(F_2) = F_2 l_2$ is a straight line and the moment $M_1$ is a horizontal line. The lever balances where the line crosses the horizontal.
The longer the arm $l_2$ , the steeper the line $M_2(F_2)$ and the smaller the force $F_2$ needed for balance: a gain in force is a loss in distance.
Inclined plane: the graphs of work $A(s)$ are straight lines from the origin. The slope of the upper line (spent work) is the force $F$ , of the lower one (useful work) is $mgh/l$ .
The lines coincide if there is no friction ( $\eta = 100\%$ ). The greater the friction force, the farther apart the lines and the lower the efficiency — the ratio of the lower line to the upper one.

Loading…

Terms

Studied 0 of 10

Formulas

Rule of the lever's equilibrium

F_1 l_1 = F_2 l_2

Quantities
Symbol	Name	Unit
$F_1$	the first force acting on the lever	N
$l_1$	arm of the first force	m
$F_2$	the second force acting on the lever	N
$l_2$	arm of the second force	m

Solved for another quantity:

F_2 = \dfrac{F_1 l_1}{l_2}

\dfrac{F_1}{F_2} = \dfrac{l_2}{l_1}

l_2 = \dfrac{F_1 l_1}{F_2}

Moment of a force

M = F\cdot l

Quantities
Symbol	Name	Unit
$M$	moment of the force	N·m
$F$	the force that turns the body	N
$l$	arm of the force	m

Solved for another quantity:

F = \dfrac{M}{l}

l = \dfrac{M}{F}

Efficiency of a machine

\eta = \dfrac{A_\text{полезн}}{A_\text{затр}}\cdot 100\%

Quantities
Symbol	Name	Unit
$\eta$	efficiency (in per cent)	%
$A_\text{полезн}$	useful work	J
$A_\text{затр}$	spent (total) work	J

Solved for another quantity:

A_\text{полезн} = \dfrac{\eta\, A_\text{затр}}{100\%}

A_\text{затр} = \dfrac{A_\text{полезн}}{\eta}\cdot 100\%

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A lever with two forces

The forces are perpendicular to the lever, so an arm is the distance from the support to the point where the force is applied. The weight of the lever itself is ignored.

The lever is in equilibrium: $M_1 = M_2$ .

Graphs

M(F_2)

Blue line — $M_2 = F_2 l_2$ , yellow horizontal — $M_1$

M_1

30 N·m

M_2

30 N·m

F_2\ (\text{to balance})

10 N

Formulas with your numbers

M_1 = F_1 l_1 = 20\cdot 1.5 = 30\ \text{N·m}

M_2 = F_2 l_2 = 10\cdot 3 = 30\ \text{N·m}

F_2 = \dfrac{F_1 l_1}{l_2} = \dfrac{20\cdot 1.5}{3} = 10\ \text{N}

How to read the graphs

Lever: the graph of the moment

M_2(F_2) = F_2 l_2

is a straight line and the moment

M_1

is a horizontal line. The lever balances where the line crosses the horizontal.

The longer the arm

l_2

, the steeper the line

M_2(F_2)

and the smaller the force

F_2

needed for balance: a gain in force is a loss in distance.

Inclined plane: the graphs of work

A(s)

are straight lines from the origin. The slope of the upper line (spent work) is the force

F

, of the lower one (useful work) is

mgh/l

The lines coincide if there is no friction (

\eta = 100\%

). The greater the friction force, the farther apart the lines and the lower the efficiency — the ratio of the lower line to the upper one.