A block with mass of 500kg is being pulled by a cable that i

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A block with mass of 500kg is being pulled by a cable that is set an angle of 12 degrees above the horizontal.If the coefficient of static friction is 1.42 and the coefficient of kinetic friction is 0.76,what is the tension in the cable just before the block starts moving and what is the tension in the cable if after the block begins to move it is kept mocing at a constant velocity?
鄙人的模糊翻译(英文为标准):
在一个区段,500 公斤的块被被设定在水平线上面的一个 12 度的角度的电缆拉着.如果静态摩擦的系数是 1.42 ,而且运动摩擦的系数是 0.76,在在区段移动之前,电缆的拉力是多少?在持续的速度移动的时候,电缆的又是多少?

兄弟的翻译比较模糊哈.还有一个问题,你的拼写有错误.自己查吧.
我来翻译一下.
一个质量为500KG的物体.受到一个和水平方向成12度角的绳子牵引.如果静态摩擦的系数是 1.42 ,而滑动摩擦的系数是 0.76,那么,请问使物体发生移动时绳子的张力是多大?又当物体已经移动后,如果要使物体保持匀速运动,绳子的张力是多大.
设绳子的张力为F,
在运动之前的力为：拉力F,分解力F1=F*sin12竖直往上,F2=F*cos12水平方向；静摩擦力f=Nμ=F2,N=Mg-F1
F*cos12=μ*(Mg-F*sin12)——F=Mgμ/(cos12+μsin12)——μ=1.42
解得：F = 500*9.8*1.42/(cos12+1.42*sin12)
=5464N
在匀速运动之后的力为：拉力F',分解力F1'=F'*sin12竖直往上,F2'=F'*cos12水平方向；滑动摩擦力f'=N'μ'=F2',N'=Mg-F1'
F'*cos12=μ'*(Mg-F*sin12)——F'=Mgμ'/(cos12+μ'sin12)——μ’=0.76
解得:F = F = 500*9.8*0.76/(cos12+0.76*sin12)
=3278N

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