關於力與運動的問題 (3)

1. (a) Use: s = ut + (1/2)at^2
2 = (1/2)a(4^2)
a = 0.25 m/s^2
Force = 1 x 0.25 N = 0.25 N
(b) Tilt the runway until the cart begins to slide.
(c) No. There is frictional force and weight component that will slow down the cart.

2. Let Ta and Tb be the tension on the strings
hence, Ta + Tb = 20g + 100
Tb = 300 - Ta (take g = 10 m/s^2)
Take moment about B,
Ta.(2) = 20g.(1.5) + 100 x 1
Ta = 200 N
hence, Tb = 100 N

3. What is the distance between the cord and the end of the bridge? This needs to be known.

4. Net force = (6 - 1gsin(10)) N = 4.264 N
Acceleration = 4.264/1 m/s^2 = 4.264 m/s^2
The answer is option B

5. P+Q = 50
Q = 50 - P
Take moment about the suspension point at Q
0.5P = 50 x 0.2
Hence, P = 20 N
Q = (50 - 20) N = 30 N
The answer is option B

6. Let T be the tension on the rod
T.cos(20) = 0.7g
T.sin(20) = 0.7a where a is the acceleration of the cart
Hence, sin(20)/cos(20) = a/g
a = 10 x tan(20) m/s^2 = 3.64 m/s^2

天同:
3. The distance between the cord and the end of the bridge is 8 m.
那麼這題應該怎樣做?