A balanced lever has two weights on it, the first with mass $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $7 m$ $7 m$ 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 $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?

1 Answer

Impact of this question

2239 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-04T22:05:42

the second weight is 1.96 m far from the fulcrum

Explanation:

Let's consider the system made up by the lever and the weights.
The system is in equilibrium.
Since we are considering only the rotative motion of the lever, the problem could be solved thanks to the 2nd principle of dynamics in its angular form:

$\sum τ = I α$ $sum tau=Ialpha$

Where $\sum τ$ $sum tau$ is the total moment of the forces acting on the system (calculated about the same point), $I$ $I$ the moment of Inertia of the lever, $α$ $alpha$ the angular acceleration of the fulcrum.
The system is in equilibrium, so $α = 0$ $alpha=0$ (in fact, the lever remains still, it does not rotate). We have:

$\sum τ = 0$ $sumtau=0$

We have 2 forces acting on the system:

$W_{1} = 7 k g \cdot g \approx 68.67 N$ $W_1= 7 kg * g ~~ 68.67 N$
$W_{2} = 25 k g \cdot g \approx 245.25 N$ $W_2= 25 kg * g ~~ 245.25 N$

The moment $τ$ $tau$ of a force $F$ $F$ about a point O is defined as:
$¯ τ = \to F \land \to d$ $bar tau= vec F ^^ vec d$
(being d the distance between O and the point where F is applied)
We have to calculate the moments of $W_{1}$ $W_1$ and $W 1$ $W1$ about the fulcrum. We have $d_{1} = 7 m$ $d_1=7m$ , we are looking for $d_{2}$ $d_2$ .
The angle between W and d is 90°: the cross products becomes simple products. Besides, the two moments must have opposite signs (we are assuming the fulcrum is located between the two weights).

$\sum τ = W_{1} \cdot d_{1} - W_{2} \cdot d_{2} = 0$ $sum tau= W_1 *d_1 - W_2*d_2=0$

$d_{2} = \frac{W_{1} \cdot d_{1}}{W_{2}} = 1.96 m$ $d_2= ( W_1 * d_1) /W_2 = 1.96 m$

Science

Math

Humanities

... and beyond

A balanced lever has two weights on it, the first with mass $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A balanced lever has two weights on it, the first with mass 7 kg 7kg7 kg and the second with mass 25 kg25kg25 kg. If the first weight is 7 m7m 7 m from the fulcrum, how far is the second weight from the fulcrum?

1 Answer

Explanation:

Related questions

Impact of this question

A balanced lever has two weights on it, the first with mass $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?