Buoyancy f = rho water VG =1.0×103kg/m3× 0.04m3×10n/kg = 400n,
The resistance of pulley block F 1=mAg-F floating =100kg×10n/kg-400n = 600n,
At this time, the mechanical efficiency η 1=W of the pulley block is w total = f1h1n (f1+g motion) × NH1= f1f/kloc.
Solution: g move =200N,
Answer: the gravity g of the moving pulley is 200 n.
(2) As shown in Figure (1), analyze the object and the moving pulley after water:
3 fc2 = g = g+ga = mag+200n =100kg×10n/kg+200n =1200n, then FC2=400N.
p = fc2×3v = 400n×3×0. 1m/s = 120 w
Answer: The motor power of string drawing is120 w. 。
(3) When the object A rises in the water at a uniform speed, the walking device, the movable pulley M and the object A are taken as the research objects, and the force analysis diagram is shown in Figure 2, the force analysis diagram of points C and D on the lever is shown in Figure 3, and the force analysis diagram of the counterweight T is shown in Figure 4.
FC 1=G-F floating
G = mg+G+magn 1 = Meg-FD 1f ' d 1？ OD=F'C 1？ commander
FC 1=F'C 1，FD 1=F'D 1
de:n 1 = 3000n- 1000 nmoc
When the object A rises at a uniform speed after leaving the water surface, the walking device, the moving pulley M and the object A are taken as the research objects. The force analysis diagram is shown in fig. 5, the force analysis diagram of counterweight E is shown in fig. 7, and the force analysis diagram of levers C and D is shown in fig. 6.
FC2=GN2=mEg-FD2F'D2？ OD=F'C2？ commander
FC2=F'C2，FD2=F'D2
N2=3000N- 1400NmOC
N 1:N2=5: 1
Solution: OC = 2m.
A: The distance of OC is 2m.