受力分析和計算(摩擦力和拉力)

(1) Let T be the tension in the string.

Take the man and wood as a single system. Because the frictional forces between man and wood are internal forces, we have, for the system moving at constant speed,
2T = Fw where Fw is the friction between wood and ground, which acts towards the left

But Fw = 0.2 x (400 + 600) N
i.e. Fw = 200 N
Hence, 2T = 200
T = 100 N

(2) As above, the frictional force between the wood A and ground = 200 N
But frictional force acting on man Fm = T =100 N
Because Fm is in direction towards the left, the frictional force acting on wood A by the man = 100 N towards the right

Therefore, resultant frictional force on wood A
= (200 - 100) N = 100 N towards the left



2013-03-06 15:52:06 補充：
Note that the frictional force between the man and wood A is 100 N, which is smaller then the max frictional force ( = 0.2 x 600 N = 120 N). This happens because the man is at rest relative to wood A. There is no sliding motion between the two.

2013-03-06 15:57:30 補充：
I wonder if the answer provided by 麻辣 is physically possible. The string tension of 120 N (towards the right) that acts on wood A is larger than the frictional force of 80N (towards the left). Wood A should then have an acceleration, instead of a constant speed.

如圖人重w1=600N，木塊A重w2=400N，人與A、A與間的動摩擦係數均為f=0.2，現人用水平力拉繩，使他與木塊一起向右做勻速直線運動，滑輪摩擦不計，求：(1)人對繩的拉力摩擦力: F1=f*w1=0.2*600=120(N)人對繩的拉力: T1=F1=120(N)
(2)A所受的摩擦力方向和大小。摩擦力: F2=f*(w1+w2)=0.2*1000=200(N)向左


2013-03-06 15:27:26 補充：
(2)漏掉人的摩擦.修正:

摩擦力: F2=f*(w1+w2-w1)=0.2*400=80(N)向左