Let the force at point E be FE and the gravity at point D be GD.
According to the stress relationship of lever T 1*5=FE*2, that is, T 1=2FE/5.
At this time, the pressure of the counterweight to the ground PS = 6000 * 5 * 10-2 = 300 N, and the tension fa1= ga-PS =1200-300 = 900 n.
According to the stress relationship of pulley: FE= 1/2* (FA 1+GD)
Therefore: t1= 2 * (1/2 * (fa1+GD))/5 = (fa1+GD)/5.
And T 1+F 1=600, so FI=600-T 1.
de:f 1 = 600-(fa 1+GD)/5 = 600-(900+GD)/5
Case 2: counterweight pressure: 4000 * 5 * 10-2 = 200 N, FA2 =1200-200 =1000 n.
Similarly:
T2 = 2 *( 1/2 *(FA2+ Guangdong)) /5 =(FA2+ Guangdong) /5
F2 = 600-(FA2+ items) /5 = 600-( 1000+ items) /5
Now there are three equations:
F 1 = 600-(900+ items) /5
F2 = 600-( 1000+ items) /5
F 1:F2=20: 19
Solve it:
F 1 = 400 Newton
F2 = 380N Newton
GD= 100N
Brought in: T2 = (Fa2+GD)/5 = (1000+100)/5 = 220n.
Answer: (1) tension fa2 =1000n; (2) Tension T2 = 220 N; (3) The gravity on the moving pulley D is 100N.
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