Question

what mass would you hang on the right side of the system in the figure shown to balance it?
(figure of a lever with 50 g mass 18 cm from fulcrum, unknown mass 12 cm from fulcrum on the other side)
options: 120 g, 50 g, 75 g, 33 g

Explanation:

Step1: Recall the principle of moments

For a lever to balance, the clockwise moment equals the counter - clockwise moment. The formula for moment is \(M = F\times d\), where \(F\) is the force (in this case, weight \(W = mg\), and since \(g\) is constant, we can use mass \(m\) instead of weight for the ratio) and \(d\) is the distance from the pivot. So, \(m_1\times d_1=m_2\times d_2\), where \(m_1 = 50\space g\), \(d_1=18\space cm\), \(d_2 = 12\space cm\) and \(m_2\) is the unknown mass.

Step2: Solve for \(m_2\)

From \(m_1\times d_1=m_2\times d_2\), we can re - arrange the formula to \(m_2=\frac{m_1\times d_1}{d_2}\). Substitute \(m_1 = 50\space g\), \(d_1 = 18\space cm\) and \(d_2=12\space cm\) into the formula: \(m_2=\frac{50\times18}{12}\). First, calculate \(50\times18 = 900\), then divide by 12: \(\frac{900}{12}=75\space g\).

Answer:

75 g