An object of mass m = 6.05 kg has an acceleration a = (1.17 m/s2)x + (-0.664 m/s2)y.Three forces act on this object:F1,F2,and F3.Given that F1 = (2.96 N)X and F2 = (-1.55 N)x + (2.05 N)y,find F3.F3=( )X + ( )Y?
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/31 08:40:39
An object of mass m = 6.05 kg has an acceleration a = (1.17 m/s2)x + (-0.664 m/s2)y.Three forces act on this object:F1,F2,and F3.Given that F1 = (2.96 N)X and F2 = (-1.55 N)x + (2.05 N)y,find F3.F3=( )X + ( )Y?
An object of mass m = 6.05 kg has an acceleration a = (1.17 m/s2)x + (-0.664 m/s2)y.Three forces act on this object:F1,F2,and F3.Given that F1 = (2.96 N)X and F2 = (-1.55 N)x + (2.05 N)y,find F3.
F3=( )X + ( )Y?
An object of mass m = 6.05 kg has an acceleration a = (1.17 m/s2)x + (-0.664 m/s2)y.Three forces act on this object:F1,F2,and F3.Given that F1 = (2.96 N)X and F2 = (-1.55 N)x + (2.05 N)y,find F3.F3=( )X + ( )Y?
力的分解,由牛顿运动定律,在x方向上F1+F2+F3=ma（x）解得F3在x方向上是5.6685.同理在y方向上F1+F2+F3=ma（y）解得F3在y方向上是-6.0672
物体质量为6.05千克,加速度为
a = 1.17x + (-0.664)y
那么作用在其上的合力为
F = ma = 7.0785x + (-4.0172)y
而已知的三个力的合力为
F1 + F2 + F3
=2.96x + [(-1.55)x + (2.05)y] + [(m)x + (n)y]
=(2.96-1.55+m)x + (...
全部展开
物体质量为6.05千克,加速度为
a = 1.17x + (-0.664)y
那么作用在其上的合力为
F = ma = 7.0785x + (-4.0172)y
而已知的三个力的合力为
F1 + F2 + F3
=2.96x + [(-1.55)x + (2.05)y] + [(m)x + (n)y]
=(2.96-1.55+m)x + (2.05+n)y
=(1.41+m)x + (2.05+n)y
有F=F1+F2+F3
得
1.41+m=7.0785
2.05+n=-4.0172
解得
m=5.6685
n=-6.0672
故F3=5.6685x+(-6.0672)y
收起
昏，这是三个力共同作用啊，你可以画图出来分析。 在x方向上，物体加速度ax=1.17m/s²，x方向受力为：F1+F2x+F3x【括号里的是下标】，也就是说：F1+F(2x)+F(3x)=max，2.96N+(-1.55N)+F(3x)=6.05Kg×1.17m/s²，所以F(3x)=5.6685N。 同理，y方向上也有F(2y)+F(3y)=may，即：2.05N+F(3y)=6.05Kg×(-0.664m/s²)，F(3y)=-6.0672N。 所以，F3=5.6685N(x)-6.0672N(y)。可以用合力表示：F3=√{[5.6685N(x)]²+[6.0672N(y)]²}≈8.3031806N，方向是第四象限，角度θ≈arctan(-1.0703361)=46.9457°