答案
已知:OB=0.9m,AB=0.6m,G=500N,h=0.4m,n=30,t=1min
求:OA=?OB=?F,W,P
解:(1)动力臂OA=OB+AB=1.5m、阻力臂OB=0.9m.
(2)F
1L
1=F
2L
2F
1×1.5m=500N×0.9m
F
1=300N
(3)W=FS=300N×0.4m=120J
(4)
P===60W
答:动力臂是1.5m,阻力臂是0.9m,双手对地面的压力是300N,身体撑起一次做的功是120J,1min内的功率是60W.
已知:OB=0.9m,AB=0.6m,G=500N,h=0.4m,n=30,t=1min
求:OA=?OB=?F,W,P
解:(1)动力臂OA=OB+AB=1.5m、阻力臂OB=0.9m.
(2)F
1L
1=F
2L
2F
1×1.5m=500N×0.9m
F
1=300N
(3)W=FS=300N×0.4m=120J
(4)
P===60W
答:动力臂是1.5m,阻力臂是0.9m,双手对地面的压力是300N,身体撑起一次做的功是120J,1min内的功率是60W.