关于加速度、牛顿运动定律An object of mass m1 hangs from a string that pa
关于加速度、牛顿运动定律
An object of mass m1 hangs from a string that passes over a very light pulley P1 as shown in the figure.The string connects to a second very light pulley P2.A second string passes around this pulley with one end attached to a wall and the other to an object of mass m2 on a frictionless,horizontal table.(a )If a1 and a2 are the accelerations of m1 and m2,respectively,what is the relationship between these accelerations?Find expressions for(b)the tensions in the strings and(c)the accelerations a1 and a2 in terms of the masses m1 and m2,and g.
An object of mass m1 hangs from a string that passes over a very light pulley P1 as shown in the figure.The string connects to a second very light pulley P2.A second string passes around this pulley with one end attached to a wall and the other to an object of mass m2 on a frictionless,horizontal table.(a )If a1 and a2 are the accelerations of m1 and m2,respectively,what is the relationship between these accelerations?Find expressions for(b)the tensions in the strings and(c)the accelerations a1 and a2 in terms of the masses m1 and m2,and g.
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(1)p1是定滑轮,一根绳上的各质点速度大小相同,p2是动滑轮,故有v2:v1=1:2恒成立.又初速度为零,故a1:a2=2:1
(2)(3)对m2受力分析有：t1=2t2=2m2a2
对m1受力分析有：m1g-t1=m1a1
故t1=(m1*m2*g)/(m1+m2)
t2=(m1*m2*g)/2(m1+m2)
a1==(m1*g)/(m1+m2)
a2==(m1*g)/2(m1+m2)
so easy
(2)(3)对m2受力分析有：t1=2t2=2m2a2
对m1受力分析有：m1g-t1=m1a1
故t1=(m1*m2*g)/(m1+m2)
t2=(m1*m2*g)/2(m1+m2)
a1==(m1*g)/(m1+m2)
a2==(m1*g)/2(m1+m2)
so easy
1年前 追问
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你说得对，我打错了，后面按错的算的a1:a2=1:2. t1=(m1*m2*g)/(m1+4m2) t2=2(m1*m2*g)/(m1+4m2) a1==(m1*g)/(m1+4m2) a2==2(m1*g)/(m1+4m2) sorry
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你能帮帮他们吗