如图1所示,一轻绳跨过一固定的滑轮,两端各连接质量分别为m1和m

如图1所示,一轻绳跨过一固定的滑轮,两端各连接质量分别为m和m的物体.为不失一般性,设m<m.放手后m将加速下降,m加速上升,加速度大小均为a.注意此时轻绳也将做变速运动,为判断绳中张力大小是否处处相等,如图1中所示可以隔离质量为Δm的一小段绳子加以研究,其受力分析如图2所示,Δmg为绳子本身所受的重力,F、F分别为上、下绳子的拉力.由牛顿第二定律有Δmg+F-F=Δma.由于整段绳为轻绳即质量可以忽略不计,故Δm趋近于零,而加速度a为有限值,因此F=F,即不论轻绳是否平衡均满足张力大小处处相等.
力学问题的研究对象中有很多所谓的轻小物体,其特点之一就是可以忽略质量,同时也可以忽略重力.从上面的分析知可同样对其作受力分析,利用牛顿运动定律求解有关问题.
例2   如图3所示,一小球在纸面内来回摆动.当轻绳OA与OB拉力相等时,摆线OC与竖直面夹角θ为(     )
A.15°   B.30°   C.45°   D.60°

答案


A


【试题分析】

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