✔ 最佳答案
13. (a) First, find the coordinates of the centre of mass, (Xo, Yo).
Take moment about the y-axis: 0.6 x (-0.5) = (0.6+0.4).(Xo)
i.e. Xo = -0.3
Take moment about the x-axis: 0.4 x (-0.1) = (0.6+0.4).(Yo)
i.e. Yo = -0.04
Use: net force = mass x acceleration
0.4g = (0.4+0.6)a, where a is the acceleration of the system, and gis the acceleration due to gravity(= 9.81 m/s^2)
hence, a = 3.924 m/s^2
Acceleration of centre of mass in x-direction = 3.924 x 0.3/0.5 m/s^2 = 2.35 m/s2
Acceleration of centre of mass in -y direction = 3.924 x (0.04/0.1) m/s2 = 1.57 m/s2
Therefore, acceleration of centre of mass = (2.35i - 1.57j)
(b) Since velocity = acceleration x time
hence, velocity v = (2.35i - 1.57j).t
(c) distance travelled = (1/2)at^2 (because s = ut + (1/2)at^2)
Hence, the coordinates (x,y) of the centre of mass is:
x = -0.5 +1.175t^2 and y = -0.785t^2
(d) Since both displacements in the x and y directions vary with t^2 (or both velocities in the x and y directions are proportional to t), the path is a straight line.
angle below x-axis = arc-tan(0.785/1.175) degrees = 33.7 degrees
17. Let d be the distance moved by the flatboat when the dog has moved 2.4 m
Because the centre of mass of the system doesn't move, hence,
4.5 x 2.4 = 18d
d = 0.6 m
Therefore, the distance of the dog from the shore
= (6.1 - 2.4 +0.6) m = 4.3 m
2011-10-16 12:55:20 補充：
Sorry, the coordinate for the centre of mass at part(d) of Q13 should be:
x = -0.3 + 1.175t^2, y = -0.04 - 0.785t^2
By combining these two equations, the path of the centre of mass is given by,
y = -0.668x - 0.24
Hence the path is a straight line.
2011-10-17 14:16:00 補充：
Q:what is 0.3 and 0.5?
A: 0.3 m is the x-corrdinate of the centr eof mass, and 0.5 m is the x-coordinate of the 0.6 kg mass. The is to transpire the acceleration of the mass to that of the "centre of mass" which represents the whole system.
