A hydraulic jack is used to move an 8896 Newton car. If the area of the output piston is 8 times that of the input piston, how much input force is required to lift the the car?
Is this where I do P=PA or F1(A2)/F2 = A1(f2)/F2
F1 = A1F2/A2
Please help set this up.
F2=8896
A1=8
F1=?
A1=?
Force = pressure times area
so for pressure p in the fluid
p A1 = F1 Input side
p A2 = F2 Output side
so
F1 /A1 = F2/A2
F1 = F2 (A1/A2)
F1 = 8896 (1/8)
Yes, you are on the right track! To solve for the input force (F1), you can use the equation:
F1 = (A1 * F2) / A2
In this equation:
- F1 is the input force (the force applied to the smaller input piston)
- A1 is the area of the input piston
- F2 is the force exerted by the car (given as 8896 Newton)
- A2 is the area of the output piston (8 times the area of the input piston)
To find the input force, we need to know the values of A1 and A2.
You are given that A1 = 8. So, plugging in the values, the equation becomes:
F1 = (8 * 8896) / A2
Now you need to find the value of A2 in order to calculate F1. Is there any additional information given about the size or area of the output piston?