A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $14 k g$ $14 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?

Question

A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $14 k g$ $14 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2016-07-31T20:12:46

$The second weight = 2 \frac{1}{7} m e t r e s$ $"The second weight "=2 1/7 metres$ from the fulcrum as an exact value

$The second weight \approx 2.143 m e t r e s$ $"The second weight "~~2.143 metres$ from the fulcrum to 3 decimal places

Explanation:

Tony B
$Assumption 1$ $color(red)("Assumption 1")$
The beam ends at the points of loading. It is not stated as such

$Assumption 2$ $color(red)("Assumption 2")$
The weight of the beam is discounted. No uniformly distributed load given.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For the beam to be in equilibrium (balanced and not moving)

All forces and moments cancel out.

$Taking moments about point B$ $color(blue)("Taking moments about point B")$

Let clockwise moments be positive
Let anticlockwise moments be negative

A moment is $force \times length of moment arm$ $"force "xx" length of moment arm"$

The force of the 'Reaction' has moment arm length of 0. So this cancels itself out giving only:

So for the system to be in equilibrium (not moving)

$(clockwise moment) + (anticlockwise moment) = 0$ $"(clockwise moment) + (anticlockwise moment)" = 0$
$↑$ $" " color(brown)(uarr )$
$We have chosen that anticlockwise rotation is negative$ $color(brown)("We have chosen that anticlockwise rotation is negative")$

$(16 \times 2) + (- 14 \times x) = 0$ $(16xx2)+(-14xx x)=0$

$32 = + 14 x$ $32=+14x$

$x = \frac{32}{14} \equiv \frac{16}{7} m e t r e s \to 2 \frac{1}{7} m e t r e s$ $x=32/14 -= 16/7 metres" " ->" " 2 1/7 metres$ as an exact value

$x \approx 2.143 m e t r e s$ $x~~2.143 metres$ to 3 decimal places

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A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $14 k g$ $14 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

A balanced lever has two weights on it, the first with mass 16kg16 kg and the second with mass 14kg14 kg. If the first weight is 2m 2 m from the fulcrum, how far is the second weight from the fulcrum?

1 Answer

Explanation:

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Impact of this question

A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $14 k g$ $14 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?