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Nana76 [90]
2 years ago
5

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2400 kg car (a large car

) resting on the slave cylinder? The master cylinder has a 1.50 cm diameter, while the slave has a 24.0 cm diameter.
Physics
1 answer:
makvit [3.9K]2 years ago
8 0

Answer:

Fm= 91.88 N

Explanation:

Pascal principle

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

The pressure is definited like this:

P=F/A

Where:

P: Pressure in pascals (Pa)

F: Force acting in the area  (N)

A  : Area where the force acts  (m²)

Pascal principle

Pm=Ps

Fm/ Am= Fs/ As  Formula (1)

Where :

Pm : Pressure on the master piston

Ps  : Pressure on the slave piston

Fm : Force on the master piston (N)

Fs:  Force on the  slave piston ((N)

Am: master piston area (m²)

As:  slave piston area  (m²)

Area Formula (A)

A= π*R²

R : piston radius

Calculation of the weight of the car (W)

W= m*g= 2400 kg*9.8m/s²= 23520 N

W = Fs

Data

Fs =  23520 N

Dm = 1.5 cm

Ds = 24 cm

Rm = 0.75 cm

Rs = 12 cm

Am = π*Rm² = π*(0.75)²

As = π*Rs² = π*(12)²

Force exerted on the master cylinder

We replace data in the formula (1)

\frac{F_{m} }{A_{m} } = \frac{F_{s} }{A_{s} }

F_{m}  = \frac{F_{s}*A_{m}  }{A_{s}}

F_{m} = \frac{(23520 N)*(\pi *(0.75)^{2})(cm^{2})}{(\pi *(12)^{2})(cm^{2})}

F_{m} = (23520 N)*\frac{(0.75)^{2} }{(12)^{2} }

Fm= 91.88 N

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Answer:

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I attached an image of the free body diagrams I drew of crate #1 and #2.  

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

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Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

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  • T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

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The normal force is equal to the x-component of the force of gravity.

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Now let's go back to this equation:

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We have 3 known variables and we can solve for the tension force.

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The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

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