【题目】宽9 m的成型玻璃以2 m/s的速度连续不断地向前行进,在切割工序处,金刚割刀的速度为10 m/s,为了使割下的玻璃板都成规定尺寸的矩形,则:
(1)金刚割刀的轨道应如何控制?
(2)切割一次的时间多长?
(3)所生产的玻璃板的规格是怎样的?
【答案】(1)割刀速度方向与玻璃板速度方向成arccos角度 (2)0.92 s (3)长9 m、宽1.84 m
【解析】(1)由题目条件知,割刀运动的速度是实际的速度,所以为合速度.其分速度的效果恰好相对玻璃垂直切割.
设割刀的速度v1的方向与玻璃板速度v2的方向之间的夹角为θ,如图所示。要保证割下均是矩形的玻璃板,则由v1是合速度得v2=v1cos θ
所以cos θ==,即θ=arccos
所以,要割下矩形板,割刀速度方向与玻璃板速度方向所成角度为θ=arccos
(2)切割一次的时间t== s≈0.92 s
(3)切割出的矩形玻璃板的规格为:长度d=9 m
宽度:l=v2t=2×0.92 m=1.84 m
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