t*y
发帖数: 2 | 1 A construction worker pulls a 5-meter plank up
the side of a building under construction by
means of a rope tied to one end of a plank.
Assume the opposite end of the plank follows a
path perpendicular to the wall of the building
and the worker pulls the rope at a rate of 0.15
meters per second. How fast is the end of the
plank sliding along the ground when it is 2.5
meters from the wall of the building? |
R*********r
发帖数: 1855 | 2 一块5米长的木板斜靠着一堵墙，墙与地面垂直，跟墙接触的一端以0.15米/秒的速度向
上运动，问木板与地面接触的一端和墙根距离2.5米时速度是多少。 |
e**y
发帖数: 54 | 3 [2.5-wdt]^2 + [(5^2 - 2.5^2)^0.5 + vdt]^2 = 5^2
where w is the horizontal and v = 0.15m/s
sovling the above equation, you can get w = (5^2 - 2.5^2)^0.5/2.5 v
【在 R*********r 的大作中提到】
: 一块5米长的木板斜靠着一堵墙，墙与地面垂直，跟墙接触的一端以0.15米/秒的速度向
: 上运动，问木板与地面接触的一端和墙根距离2.5米时速度是多少。
|
t*y
发帖数: 2 | |
G******i
发帖数: 163 | 5 Let's dig a bit deeper.
Let x be the distance from the bottom end of the plank to the wall.
Your solutions shows dx/dt = -0.15*sqrt(25-x^2)/x.
Thus, the limit speed is infinity as x approaches zero.
In reality, this certainly is impossible.
What's wrong with the above analysis?
【在 e**y 的大作中提到】
: [2.5-wdt]^2 + [(5^2 - 2.5^2)^0.5 + vdt]^2 = 5^2
: where w is the horizontal and v = 0.15m/s
: sovling the above equation, you can get w = (5^2 - 2.5^2)^0.5/2.5 v
|
l********e
发帖数: 3632 | 6 很有意思，有什么解释？
【在 G******i 的大作中提到】
: Let's dig a bit deeper.
: Let x be the distance from the bottom end of the plank to the wall.
: Your solutions shows dx/dt = -0.15*sqrt(25-x^2)/x.
: Thus, the limit speed is infinity as x approaches zero.
: In reality, this certainly is impossible.
: What's wrong with the above analysis?
|
B********e
发帖数: 10014 | 7 呵呵，难道就是传说中的鞭梢效应？
等你揭迷底啦
【在 G******i 的大作中提到】
: Let's dig a bit deeper.
: Let x be the distance from the bottom end of the plank to the wall.
: Your solutions shows dx/dt = -0.15*sqrt(25-x^2)/x.
: Thus, the limit speed is infinity as x approaches zero.
: In reality, this certainly is impossible.
: What's wrong with the above analysis?
|
l********e
发帖数: 3632 | 8 想了想，结论是这样：如果有绝对刚体，那么速度自然可以区域无穷。
但是实际上要考虑物体的内部弹性，当物体贴墙的一段匀速运动时候，实际上物体受到
的力应该是向上控制物体运动的力，还有墙对物体水平反方向的力，这样物体靠近地面
的一段在离墙很近的时候力应该很小（是接近竖直方向内部应力的水平投影）速度当然
到不了无穷了。 |
P*****J
发帖数: 1745 | 9
就算考虑到板子是绝对刚体，末端也不可能达到无穷的速度。这是因为重力加速度是有
限的。
Mumbo jumbo, doesn't make sense.
如果板子撞墙时末端还在地面上，所需要的线速度是正无穷。除非有无穷大的重力加速度，班子末端不可能在有限时间内加速到无穷快。所以板子末端在碰墙之前就已经离地了。
【在 l********e 的大作中提到】
: 想了想，结论是这样：如果有绝对刚体，那么速度自然可以区域无穷。
: 但是实际上要考虑物体的内部弹性，当物体贴墙的一段匀速运动时候，实际上物体受到
: 的力应该是向上控制物体运动的力，还有墙对物体水平反方向的力，这样物体靠近地面
: 的一段在离墙很近的时候力应该很小（是接近竖直方向内部应力的水平投影）速度当然
: 到不了无穷了。
|
G******i
发帖数: 163 | 10 Nice argument!
The next question is: When/where does the plan leave the ground?
速度，班子末端不可能在有限时间内加速到无穷快。所以板子末端在碰墙之前就已经离
地了。
【在 P*****J 的大作中提到】
:
: 就算考虑到板子是绝对刚体，末端也不可能达到无穷的速度。这是因为重力加速度是有
: 限的。
: Mumbo jumbo, doesn't make sense.
: 如果板子撞墙时末端还在地面上，所需要的线速度是正无穷。除非有无穷大的重力加速度，班子末端不可能在有限时间内加速到无穷快。所以板子末端在碰墙之前就已经离地了。
|
P*****J
发帖数: 1745 | 11 This is not hard. The ground end of the plank leaves the ground when the
normal force at that end reaches zero. According to my derivation, it occurs
when the following equation is satisfied:
3/2*g/L*cos(theta) = v^2/L^2 *tan(theta)*sec^2(theta)
where theta is the angle between the plank and the ground, v is the velocity
of the wall end, g is the gravity, and L is the length of the plank. The
quantity to the left is the angular acceleration of the plank due to the
gravity (assuming the normal fo
【在 G******i 的大作中提到】
: Nice argument!
: The next question is: When/where does the plan leave the ground?
:
: 速度，班子末端不可能在有限时间内加速到无穷快。所以板子末端在碰墙之前就已经离
: 地了。
|
