有一箱重W 置於水平地板上

Let F be the required force.
Hence, for the box to slide, we have,
F.cos(45) = (us).R ---------------- (1)
where R is the normal reaction

But R + F.sin(45) = W
i.e. R = W - F.sin(45)
Therefore, equation (1) becomes,
F.cos(45) = (us).(W - F.sin(45))
F = (us)W/[(1+us)cos(45)]

有一箱重W 置於水平地板上 箱與地面間之靜摩擦係數=Us 今以與水平成Q=45度角的方向斜向上方拉箱 使箱在地板上滑動 則所需之拉力F=至少為多少?Sol:令R=地面反力ΣFy=W-R-F*sinQ=0 => R=W-F*sinQΣFx=F*cosQ-R*Us=0 => F=R*Us/cosQ上面兩式消除R:F=(W-F*sinQ)*Us/cosQF*cosQ=W*Us-F*Us*sinQF=W*Us/(Us*sinQ+cosQ).......ans