手柄AB長0.25m,在柄端B處作用一其大小為40N的力F,則此力對A點的最大力矩以及相應(yīng)的α角的值為()
A.mA(F)=100.0N·m,α=60° B.mA(F)=-10.0N·m,α=30° C.mA(F)=10.0N·m,α=60° D.mA(F)=8.7N·m,α=90°
圖示力F,對y軸之矩為()
A.2√2FB.-2√2FC.3√2F/2D.-3√2F/2
圖示力F,對x軸之矩為()
A.2√2F B.-2√2F C.3√2F D.-3√2F