A 50 kg block of wood is pressed against a smooth wall by a force perpendicular to the wall.If the coefficient of static friction is 0.13 ,what is the minimum force that is necessary to keep the block of wood in place?模糊的翻译:一个50 公斤

A 50 kg block of wood is pressed against a smooth wall by a force perpendicular to the wall.If the coefficient of static friction is 0.13 ,what is the minimum force that is necessary to keep the block of wood in place?模糊的翻译:一个50 公斤
A 50 kg block of wood is pressed against a smooth wall by a force perpendicular to the wall.If the coefficient of static friction is 0.13 ,what is the minimum force that is necessary to keep the block of wood in place?
模糊的翻译:
一个50 公斤木块被按没光滑的墙壁由力量垂线对墙壁.如果静态摩擦系数是0.13,以最低到什么的力量来保持木块到位?
At what altitude above the Earth's surface would the gravitational acceleration be 4.9m/s^2 The mass of the Earth is 5.98 x 10^24 kg and radius is 6.37 x 10^6 m
模糊翻译:
在什么高度在地球之上表面重心加速度会是4.9m/s^2?地球的重量是5.98 x 10^24 公斤并且半径是6.37 x 10^6 m.

A 50 kg block of wood is pressed against a smooth wall by a force perpendicular to the wall.If the coefficient of static friction is 0.13 ,what is the minimum force that is necessary to keep the block of wood in place?模糊的翻译:一个50 公斤
第一题:
50Kg的木块被压在光滑墙壁上,压力和墙面垂直,静摩擦系数为0.13,为了让木块保持静止,至少要多大的压力.
因为是光滑墙壁,所以只有手和木块的接触面上有静摩擦,
要使木块不滑动,摩擦力就不能小于重力,所以最小摩擦力
等于重力,于是
F x 0.13 =50g
压力F便可求得
第二题：
在多高的时候,重力加速度的数值是4.
重力其实是万有引力的一个分力,但根据本题题意,不考虑
地球自转的情况下,重力就近似于万有引力.
设地球的质量为M,半径是R;在高度h处,有一质量为m的物体,
且此高度处的重力加速度g=4.5
则,
GMm/(R+h)^2 = mg
(左边是万有引力,右边是重力)
把m消掉就可以求得高度h了.

楼上的都是对的：
我来帮你解出结果：
(1) μF = mg
F = mg /μ = 50*9.8/0.13 = 3769N
(2) GMm/(R+h)^2 = mg'
h = sqrt(GM/g') -R = sqrt[6.67259×10^(-11)*5.98 x 10^24/4.9]-6.37*10^6
=2.65*10^6m
还有一个...

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楼上的都是对的：
我来帮你解出结果：
(1) μF = mg
F = mg /μ = 50*9.8/0.13 = 3769N
(2) GMm/(R+h)^2 = mg'
h = sqrt(GM/g') -R = sqrt[6.67259×10^(-11)*5.98 x 10^24/4.9]-6.37*10^6
=2.65*10^6m
还有一个简便解法
因为万有引力与距离的平方成反比
所以 r/R = sqrt(g/g') g=9.8m/s^2
r = 1.414R
h = r-R = 0.414R = 6.37*10^6*0.414 = 2.64*10^6m

收起

不对，是千克不是克，
摩擦力f=Mg=500=Nμ=N*0.13
N为物体受到正压力=500/0.13=3846.15牛
题目的意思是重力加速度减少一半：
根据他的公式：
GMm/[(R+h)^2] = mg/2= (GMm/[(R)^2])/2
2/[(R+h)^2]=1/[(R)^2]
2*[(R)^2]=(R+h)^2
(根号2)*R=R+h
h=[(根号2)-1]*R