题目内容
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120158447518.jpg)
(1)当升降机处于静止时,求弹簧的弹力和斜面以对物体的支持力分别为多大?
(2)当升降机以加速度2m/s2加速下降时,求弹簧的弹力和斜面以对物体的支持力分别为多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241201584471455.gif)
(1)20N 20
N(2)16N 16
N
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120158463477.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120158463477.jpg)
(1)当升降机静止时,对物体受力分析,并建立坐标系如图![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241201584942191.gif)
由平衡条件得:
F=Gx-----------------------(1) 1分
FN=Gy----------------------(2) 1分
F=Gsinθ
FN=Gcosθ
则:弹簧的拉力为:F="20N " 1分
斜面对物体的支持力为:FN=20
N 1分
(2)当升降机加速下降时对物体受力分析,并建立坐标系如图所示
由牛顿第二定律得:
G-FNy-Fy="ma------------------(1) " 1分
Fx=FNX-------------------------(2) 1分
G-FNcosθ-Fsinθ=ma
Fcosθ=FNsinθ
解得:F="16N " 2分
FN=16
N 2分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241201584942191.gif)
由平衡条件得:
F=Gx-----------------------(1) 1分
FN=Gy----------------------(2) 1分
F=Gsinθ
FN=Gcosθ
则:弹簧的拉力为:F="20N " 1分
斜面对物体的支持力为:FN=20
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120158463477.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241201585253270.gif)
由牛顿第二定律得:
G-FNy-Fy="ma------------------(1) " 1分
Fx=FNX-------------------------(2) 1分
G-FNcosθ-Fsinθ=ma
Fcosθ=FNsinθ
解得:F="16N " 2分
FN=16
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120158463477.jpg)
![](http://thumb2018.1010pic.com/images/loading.gif)
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