Problem 3.45

พิจารณา F.B.D. ของมวล m

จาก \(ΣFx_{y}= ma_{y}\)

y

\(f = 2\mu nn - mgcos\theta = m(0)\)

\(n = mgcos\theta _{\theta}\)

\(mgsin\theta\)

\(Tmgcos\theta\)

n

mg

จาก \(ΣF_{x}= ma_{x}\)

\(2\mu n - T - mgsin\theta = m(-a)\)

\(2\mu mgcos\theta - T - mgsin\theta = -ma \cdots \cdots (1)\)

พิจารณา F.B.D. ของมวล \(2m\)

จาก \(ΣFx_{y}= ma_{y}\)

yT

\(f = \mu N\)

\(N - 2mgcos\theta = 2m(0)\)

\(2mgsin\theta N = 2mgcos\theta ^{\theta}\)

\(2mgcos\theta\)

N

\(2mg\)

จาก \(ΣF_{x}= ma_{x}\)

\(\mu N + T - 2mgsin\theta = 2m(-a)\)

\(\mu (2mgcos\theta ) + T - 2mgsin\theta = -2ma\)

\(2\mu mgcos\theta + T - 2mgsin\theta = -2ma\cdots \cdots (2)\)

\((1) + (2) ; 4\mu mgcos\theta - 3mgsin\theta = -3ma\)

\(a = gsin\theta -\frac{4\mu gcos\theta}{3}\)

a แทนใน \((1)\) จะได้

\(2\mu mgcos\theta - T - mgsin\theta = -m(gsin\theta -\frac{4\mu gcos\theta}{3})\)

\(T =\frac{2\mu mgcos\theta}{3}\;\blacksquare\)