Question

a horizontal meter stick has a mass of 245 g. three weights ride on the meter stick: 265 g at 43.3 cm, 187 g at 76.5 cm, and 199 g at 98.1 cm.
at what location x on the meter stick would the system be in balance if it were suspended from that point?

Explanation:

Step1: Calculate the total mass

The total mass \(M\) is the sum of the mass of the meter - stick \(m_{stick}\) and the masses of the weights \(m_1\), \(m_2\), \(m_3\).
\(M=m_{stick}+m_1 + m_2+m_3\)
\(m_{stick}=245g\), \(m_1 = 265g\), \(m_2=187g\), \(m_3 = 199g\)
\(M=245 + 265+187 + 199=906g\)

Step2: Calculate the moment - weighted sum

The center of mass of the meter - stick is at \(x_{stick}=50cm\) (since for a uniform rod, the center of mass is at its mid - point).
The moment - weighted sum \(S\) is given by \(S=m_{stick}x_{stick}+m_1x_1+m_2x_2+m_3x_3\)
\(x_1 = 43.3cm\), \(x_2=76.5cm\), \(x_3 = 98.1cm\)
\(S=245\times50+265\times43.3+187\times76.5+199\times98.1\)

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Step3: Calculate the balance point \(x\)

Using the formula for the center of mass \(x=\frac{S}{M}\)
Since \(S = 57551.9\) and \(M = 906\)
\(x=\frac{57551.9}{906}\approx63.5cm\)

Answer:

\(63.5\)