- #1
brotherbobby
- 618
- 152
- Homework Statement
- Rank in order, from largest to smallest, the magnitudes of the forces, ##\vec{F}_a, \vec{F}_b## and ##\vec{F}_c## required to balance the masses (in kilograms) shown in the diagram below.
- Relevant Equations
- Pascal's law states that for a pair of cylinders of areas ##A_1## and ##A_2## "connected" over a fluid body and at the same height, the forces required to balance them are in direct proportion to the areas : ##\frac{F_1}{F_2} = \frac{A_1}{A_2}##.
Additionally, the pressure at a point in a fluid a distance ##d## below the surface is ##p = p_0 + \rho g d##, where ##p_0## is the pressure at the surface of the fluid which is usually atmospheric pressure : ##p_0 = p_{\text{atm}}## (usually)
Answer : Using Pascal's law, this is my answer : ##\color{blue}{\boxed{\vec F_a = \vec F_c < \vec F_b}}##.
Reasoning :
Forces ##F_a## and ##F_c## are equal because the pressures required at the two cylinders for case (c) is the same as that required in (a). It doesn't matter how many of those cylinders of liquid exist for the case (c) or (a). For as long as the masses (and therefore weights) on them remain the same, the same force would be necessary to keep the brakes (pistons) in place.
However, force ##F_a < F_b## using Pascal's law as stated above. The pressure to be generated by the force at the second cylinder in (b) is more than the same for (a) since a 800 kg mass is heavier than a 500 kg mass. This excess pressure at the second cylinder for (b) must be the same at the same level for the first cylinder. Hence the force exerted at the first cylinder has to be correspondingly greater at (b) than at (a).
Numerically, ##F_b = 1.6 F_a## since ##\frac{8}{5} = 1.6##.
Is my answer correct?