取AB桿為自由體如圖所示,W、RA、NB三力平衡必相交於一點 O
μ = tan Φ = 0.6/1.6 = 3/8 =0.375
另外解法
NA=400(N)
ΣMA=0
NB×1.6-400×0.6 = 0
NB = 150(N)
fs = μ×NA = NB
μ×400 = 150
μ = 150 / 400 = 0.375