Question

已知作用在A點的三個力F₁=(3,4),F₂=(2,-5),F₃=(3,1)且A(1,1),則合力F=F₁+F₂+F₃的終點坐標為（    ）

A.(9,1)             B.(1,9)               C.(9,0)                D.(0,9)

Answer 1

解析:設點B(3,4),C(2,-5),D(3,1),

F₁==(3,4),F₂==(2,-5),F₃==(3,1),

∴F=F₁+F₂+F₃=(3,4)+(2,-5)+(3,1)=(8,0).

設F終點為(a,b),

又(a,b)-(1,1)=(8,0),∴(a,b)=(9,1).

∴合力F的終點坐標為(9,1).

答案:A