Found: (1)η=? (2)ρ=?
Solution:
(1) a. When lifting armor with a moving pulley,
① Take the object A and the moving pulley as the research objects, and the force analysis is as shown in figure 1:
② Take the crown block as the research object, and the stress analysis is shown in Figure 2:
(3) Force analysis of lever point A and point B is shown in Figure 3:
④ The force analysis of counterweight B is shown in Figure 4:
From top to bottom, 2t 1 = Gjia +G0? ①
FA 1=G0+3T 1？ ②
f’a 1？ lOA = F’b 1？ lOB？ ③
N 1+FB 1=G B ④
Substitution value
n 1 = p 1S = 3.5× 104 pa×0.02 m2 = 700n
T 1 = T’ 1，fa 1 = F’a 1，FB 1 = F’b 1，
G2-G0 = 1300n；
B. When the object C is immersed in water and rises at a uniform speed,
⑤ Taking the object C and the moving pulley as the research objects, the force analysis is shown in Figure 5:
⑥ Taking the crown block as the research object, the stress analysis is shown in Figure 6:
⑦ Force analysis of lever point A and point B is shown in Figure 7:
Today, the force analysis of counterweight B is shown in Figure 8:
Start from scratch? FA2=G0+3T2？ ⑤
f’A2？ Loa = B2? lOB？ ⑥
N2+FB2=G B？ ⑦
The substitution value N2 = P2S = 5.6×104pa× 0.02m2 =1120n.
T2 = T′2，FA2 = F′A2，FB2 = F′B2，
T2=Pv=20W0. 1m/s=200N can be obtained from the image.
That is, 5G b -2G0= 100N.
The solutions are G0= 100N and G B = 1400N.
When lifting the object A, the mechanical efficiency of the pulley block is η=W useful W total =W useful W useful +W quantity =GhGh+G moving h=G A G A +G moving =100N100N+65438×100% ≈ 965438+.
(2) The gravity of object C is G C =5G0=5× 100N=500N.
∫2T2+F floating =G0+G C? ⑧
The buoyancy of ∴c object is f float = G0+GC-2t2 =100n+500n-2× 200n = 200n.
The volume of the object c is V C =V row =F floating ρ water g = 200 n1.0× kg/m3×10n/kg = 0.02m3,
Is the density of object c ρ=m C V C =G C V? g = 500n 0.02 m3× 10n/kg = 2.5× 103kg/m3。
Answer: (1) The mechanical efficiency of pulley block is 91%when lifting object A;
(2) The density of object C is 2.5× 103kg/m3.