Answer:
Force applied to smaller cross section is
= 82.63 N
Explanation:
As we know
[tex]F_2 A_1 = F_1 A_2[/tex]
where [tex]F1, F2[/tex] signifies the weight of the two chair in a hydraulic-lift system
And [tex]A_1, A_2[/tex] signifies the area of the two respective chairs in a hydraulic-lift system
Given -
[tex]F2=1400[/tex] N
[tex]A1 =1220[/tex] Square centimeter
[tex]A_2 = 72[/tex] Square centimeter
Substituting the given values in above equation, we get -
[tex]1400 * 72 = F1 * 1220\\F2 = 82.63[/tex]
Force applied to smaller cross section is
= 82.63 N
The magnitude of force to be applied on the smaller piston for lifting the chair is of 82.62 N.
Given data:
The weight of Chair is, W = 1400 N.
The cross - section area of larger piston is, A = 1220 cm².
The cross - section area of smaller piston is, a = 72 cm².
The given problem is based on the concepts of Pascal's law, which says that the pressure applied at any one point on the walls of vessel, gets equally distributed throughout the vessel.
Therefore,
P Â = p
Here, P is the pressure on larger section and p is the pressure on smaller piston.
So,
W/A = f/a
Solving as,
1400/1220 = f / 72
f = 1400/1220 × 72
f = 82.62 N
Thus, we can conclude that the magnitude of force to be applied on the smaller piston for lifting the chair is of 82.62 N.
Learn more about the Pascal's law here:
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