Then: Na2 = f "= P2S = 4×103pa× 5×10-2m2 = 200n.
∫G = FA2+NA2
∴fa2=g-na2= 1200n-200n= 1000n.
(2) When the vertical downward pulling force exerted by Cheng Xiao at point H is T 1, the stress of static counterweight A is shown in Figure 2:
Then: na1= f ′ = p1s = 6×103 pa× 5×10-2m2 = 300n.
∫G = fa 1+na 1
∴fa 1=g-na 1= 1200n-300n=900n.
The advantage of moving pulley d is that it saves half the effort. The pulling force of lever EH on e is f-pull1= fa1+g-move2.
The stress of Cheng Xiao itself is shown in Figure 3:
Then: T 1=G ren -f 1.
The stress analysis of lever EH is shown in Figure A in Figure 4:
Balance in horizontal position. ∴ According to the lever balance condition: T 1F pull 1 = oeoh.
Namely: g people? F1fa1+gmove2 = oeoh, and the value of the substituted data is: 600NF 1900N+G move 2 = 25-①.
Similarly, when the vertical downward pulling force exerted by Cheng Xiao at point H is T2, the stress analysis of lever EH is shown in Figure 4:
Available: g people? F2FA2+G move 2=OEOH, and the substitution data: 600NF2 1000N+G move 2 = 25-②.
According to the meaning of the question: F 1F2 = 20 19-③.
By solving the equations composed of ① ② ③, we get:
G mobile =100n; f 1 = 400n； F2=380N。
∴T2=G people -F2 = 600n-380n = 220n.
Answer: (1) tension fa2 =1000n;
(2) Tension T2 = 220 N;
(3) The gravity on the moving pulley D is g = 100N.