Can someone please help me with physics?

Question

I cannot understand this problem. Please help me :( I would appreciate it a lot. Thank you much.

Piston 1 (in the figure linked) has a diameter of 0.30 in. Piston 2 has a diammeter of 1.5 in. In the absence of friction, determine the force necessary to support the 500 lb weight.

http://www.webassign.net/sf/p9_24.gif

Answer is in pounds (lb)

Answer 1

Piston 1 (in the figure linked) has a diameter of 3.0 in. Piston 2 has a diammeter of 1.5 in. In the absence of friction, determine the force necessary to support the 500 lb weight.

Answer: The force required at piston 1 = 125 lb. This is because the pressure in piston 1 = pressure in piston 2
i.e. 125 /one quarter area = 500/ area.
Area of piston 2 = pi x [3.0/2]^2 sq. in. Area of piston 1 = pi x [1.5/2]^2 sq. in
Use the principle of moments to find force F:

ie. 125 lb x 2.0 = F x (2 + 10)
Hence F = [125 x 2] / (2 + 10)
= 125 / 6 lb force

Answer 2

I assume that the picture is a fluid pump. The pressure in the fluid will be 500lb/A1, where A1 is the area of piston 1. This same pressure will appear at piston 2. The force from the fluid at piston 2 will be its pressure times the area of piston 2, which is 500lb*A2/A1. Now you have a lever arm supplying that force. The mechanical advantage of that lever is 12in/2in; the force on the piston from the lever will then be 6*F which must equal 500lb*A2/A1

Answer 3

F x 12 = P x 2 where P is the force acting on piston 1.

P = 6F.

Pressure = force/ area.

Pressure on the piston 1 = pressure on the piston 2.

6F/ A1 = 500 / A2.

F = (A1 / A2)*(500/ 6) lb.

But A1/ A2 = (D1/D2) ^2. -------D’s are the diameters.

(D1/D2)^2 = (0.3 /1.5) ^2 = (1/5)^2 = (1/25)

F = (1/25) * (500/6) lb. = 20/6 = 3.3 lb.

Answer 4

4 lbs assuming incompressible fluid. Draw a free body diagram...EVERTIME in physics. The pressure of the 500lb object on an area of 2.25 in^2 is 222.222lbs/in^2. That is equal to the other side...equal and opposite reaction. 222.222lbs/in^2 multiplied by the other area (0.3^2 or 0.09in^2) is 20. This is 20 lbs of force needed to lift the 500lb weight. Now equalize the work done. Work=Force*Distance and W1=W2.
Soooo F1*D1=F2*D2...plug and chug...20lbs*2in= x lbs*10in. Solve.

Answer 5

well..
this problem..needs two basic concepts,,
first...the principle of hydrualics..which says...if a pressure is appiled on the surface of a fluid(liquid or gas)..it is uniformly distributed to the whole fluid....
second...the principle of equal and opposite torques balancing each other in the handle..
so,
the solution goes as follows...

since pressure =(force/cross sectional area )...equating pressure at the two openings of the tube..
we have..
(F1/A1) = (F2/A2)...
ie.,F2=F1/A1 X A2
there fore...we get...

F2= [(.3)(.3)(pi)/4] X 500/[(1.5)(1.5)(pi)/4]
F2=20 lb.

now,
this force...tries to push the handle up causing a torque in the anti clockwise direction(taking the ixed point as that where the lenghts are measured...
and our force F...produces another torque which opposes the former..
so we have...
upward torque =downward torque
ie.,
(F2)(2) = F (10+2)
so,
F=(F2)2/12
=20 X 2/12
=20/6=thats approximately...around...3.34 lb..
so..ur answer shud prolly be..

3.34 lb...
i hope u got wat u wanted..nething more write again...

(im a bit bad at math...so any calculations goin wrong...pls frogive...)