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

Jan 4, 2016

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:

$\sum \tau = I \alpha$

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

$\sum \tau = 0$

We have 2 forces acting on the system:

• ${W}_{1} = 7 k g \cdot g \approx 68.67 N$
• ${W}_{2} = 25 k g \cdot g \approx 245.25 N$

The moment $\tau$ of a force $F$ about a point O is defined as:
$\overline{\tau} = \vec{F} \wedge \vec{d}$
(being d the distance between O and the point where F is applied)
We have to calculate the moments of ${W}_{1}$ and $W 1$ about the fulcrum. We have ${d}_{1} = 7 m$, we are looking for ${d}_{2}$.
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).

$\sum \tau = {W}_{1} \cdot {d}_{1} - {W}_{2} \cdot {d}_{2} = 0$

${d}_{2} = \frac{{W}_{1} \cdot {d}_{1}}{W} _ 2 = 1.96 m$