Problem 3.55 Challenge

พิจารณา F.B.D. ของมวล m

จาก \(ΣF_{y}= ma_{y}\)

a

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

F

\(f_{1}n = mg\)

mg

จาก \(ΣF_{x}= ma_{x}\)

n

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

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

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

พิจารณา F.B.D. ของมวล M

จาก \(ΣF_{y}= ma_{y}\)

\(\frac{a}{n}\)

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

\(N = n + Mg\)

f₁

\(N = mg + Mg\)

f₂

จาก \(ΣF_{x}= ma_{x}\)

Mg

\(Nf_{1} - f_{2} = Ma\)

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

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

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

\(\mu _{1}m^{2}g - \mu _{2}m^{2}g + \mu _{2}mMg = MF - \mu _{1}mMg\)

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