Problem 4.60 Challenge

F.B.D. ของวัตถุทรงกระบอกมวล M รัศมี R

จาก \(ΣF_{x}= 0\)

F

\(f_{1} = \mu n_{1}\)

\(n_{1} - \mu n_{2} = 0\)

\(\frac{R}{_{cm}}\)

\(n_{1} = \mu n_{2} \cdots \cdots (1)n_{1}\)

Mg

จาก \(ΣF_{y}= 0\)

\(f_{2} = \mu n_{2}\)

\(F + n_{2} + \mu n_{1} - Mg = 0n_{2}\)

\(F + n_{2} + \mu n_{1} = Mg \cdots \cdots (2)\)

จาก \((1)\) และ \((2)\) จะได้ \(n_{1} = \frac{\mu (Mg - F)}{1 + \mu ^{2}}\)

\(n_{2} =\frac{Mg - F}{1 + \mu ^{2}}\)

พิจารณาจุด cm เป็นจุดหมุน จะได้ \(Σ\vec{M} = 0\)

\(FR - \mu n_{1}R - \mu n_{2}R = 0\)

\(F - \mu (\frac{\mu (Mg - F)}{1 + \mu ^{2}}) - \mu (\frac{Mg - F}{1 + \mu ^{2}}) = 0\)

\(F + \mu ^{2}F - \mu ^{2}Mg + \mu ^{2}F - \mu Mg + \mu F = 0\)

\((1 + \mu + 2\mu ^{2})F = (1 + \mu )\mu Mg\)

\(F =\frac{(1 + \mu )\mu Mg}{1 + \mu  + 2\mu ^{2}}\;\blacksquare\)