A balanced lever has two weights on it, the first with mass $3 k g$ $3 kg$ and the second with mass $2 k g$ $2 kg$ . If the first weight is $4 m$ $4 m$ from the fulcrum, how far is the second weight from the fulcrum?

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A balanced lever has two weights on it, the first with mass $3 k g$ $3 kg$ and the second with mass $2 k g$ $2 kg$ . If the first weight is $4 m$ $4 m$ from the fulcrum, how far is the second weight from the fulcrum?

3 Answers

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Answer 1 · 2018-04-17T16:24:58

The distance is $= 6 m$ $=6m$

Explanation:

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The mass $M_{1} = 3 k g$ $M_1=3kg$

The mass $M_{2} = 2 k g$ $M_2=2kg$

The distance $a = 4 m$ $a=4m$

Taking moments about the fulcrum

$M_{1} \times a = M_{2} \times b$ $M_1xxa=M_2xxb$

$b = \frac{M_{1} \times a}{M_{2}} = \frac{3 \cdot 4}{2} = 6 m$ $b=(M_1xxa)/(M_2)=(3*4)/(2)=6m$

Answer 2 · 2018-04-17T17:32:55

The second mass ( $2 k g$ $2kg$ ) is $6 m$ $6m$ from the fulcrum

Explanation:

This is a torque-balancing exercise!

Torque is an angular force (meaning it travels in a circle) that is equal to a linear force times the distance between the force and the center of rotation.

In formula-speak: $τ = F \times d$ $tau=Fxxd$ , where $τ$ $tau$ is the torque.

We have a balanced lever, which means that the torque exhibited by the right hand side is equal (and opposite!) to the torque exhibited by the left hand side:

$τ_{L} + τ_{R} = 0 \Rightarrow τ_{L} = - τ_{R}$ $tau_L+tau_R=0 rArr tau_L=-tau_R$

$F_{l} \times d_{l} = - F_{r} \times d_{r}$ $F_lxxd_l=-F_rxxd_r$

If we consider the fulcrum as the centr of rotation, then the distances (the torque or moment arms) are in opposite directions.

This handles our negative sign, since both forces are in the same direction.

Now, let's assemble the balancing equation. if Force is mass times acceleration, and we're worried about gravitational forces, then our equation looks like this:

$m_{l} \times g \times d_{l} = - m_{r} \times g \times d_{r}$ $m_lxxgxxd_l=-m_rxxgxxd_r$

We can get rid of the gravitational constants since its an equal multiplication for both sides, and plug in our masses. We also know the distance for the $3 k g$ $3kg$ body, so we'll add that in too:

$3 k g \times 4 m = - 2 k g \times d_{r}$ $3kgxx4m=-2kgxxd_r$

$\frac{3}{2} \cdot 4 m = - d_{r}$ $3/2*4m=-d_r$

$d_{r} = - 6 m$ $d_r=-6m$

Now, what we calculated is a position, which is a vector term, meaning that it has both magnitude and direction (the negative sign). What is being asked for is the distance, which is a scalar term.

Therefore our final answer will remove the negative sign:

$d_{r} = 6 m$ $color(green)(d_r=6m)$

Answer 3 · 2018-04-18T10:38:47

$6 \ "m"$

Explanation:

For a balanced lever, we have the following relationship:

$m_1d_1=m_2d_2$

$m_1,m_2$ are the masses of the two objects

$d_1,d_2$ are the distances of the two objects from the fulcrum

And so, we get:

$3 \ "kg"*4 \ "m"=2 \ "kg"*d_2$

$d_2=(3color(red)cancelcolor(black)"kg"*4 \ "m")/(2color(red)cancelcolor(black)"kg")$

$=6 \ "m"$

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A balanced lever has two weights on it, the first with mass $3 k g$ $3 kg$ and the second with mass $2 k g$ $2 kg$ . If the first weight is $4 m$ $4 m$ from the fulcrum, how far is the second weight from the fulcrum?

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A balanced lever has two weights on it, the first with mass 3 kg 3kg3 kg and the second with mass 2 kg2kg2 kg. If the first weight is 4 m4m 4 m from the fulcrum, how far is the second weight from the fulcrum?

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Explanation:

Explanation:

Explanation:

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A balanced lever has two weights on it, the first with mass $3 k g$ $3 kg$ and the second with mass $2 k g$ $2 kg$ . If the first weight is $4 m$ $4 m$ from the fulcrum, how far is the second weight from the fulcrum?