Problem 3.54 Challenge

พิจารณา F.B.D. ของมวล m

aจาก \(ΣF^{y}= ma^{y}\)

\(n - mg = m(0)\)

T

\(f_{1}n = mg\)

mg

จาก \(ΣF_{x}= ma_{x}\)

n

\(f_{1} - T = m(-a)\)

\(T - f_{1} = ma\)

\(T - \mu _{1}n = ma\)

\(T - \mu _{1}mg = ma\cdots \cdots (1)\)

พิจารณา F.B.D. ของมวล M

\(\frac{a}{n}\) จาก \(ΣF_{y}= ma_{y}\)

\(N - n - Mg = M(0)\)

\(N = n + Mg\)

\(T\frac{f_{1}}{f_{2}}F\)

\(N = mg + Mg\)

Mg

Nจาก \(ΣF_{x}= ma_{x}\)

\(F - f_{1} - f_{2} - T = Ma\)

\(F - \mu _{1}n - \mu _{2}N - T = Ma\)

\(F - \mu _{1}(mg) - \mu _{2}(mg + Mg) - T = Ma\cdots \cdots (2)\)

\((2)/(1)\) จะได้ \(\frac{F - \mu _{1}mg - \mu _{2}mg - \mu _{2}Mg - T}{T - \mu _{1}mg}=\frac{Ma}{ma}\)

\(mF - \mu _{1}m^{2}g - \mu _{2}m^{2}g - \mu _{2}mMg - mT = MT - \mu _{1}mMg\)

\(\therefore T =m(\frac{F - (\mu _{1} + \mu _{2})mg + (\mu _{1} - \mu _{2})Mg}{m + M})\;\blacksquare\)