Force and Pressure on Hydraulic System

[BlackSideburns]

Rank in order from largest to smallest, the forces required to balance the the masses (in kgs)
You can find the diagram and answer at this link on the third slide http://www.gwu.edu/~phy21bio/Presentations/PHYS1021-15a.pdf
Equation
P=F/A

The Attempt at a Solution

I'm really having trouble picturing the FBDs and seeing how this works. It's easy when comparing the two with a single weight but the one with two is what trips me up.

 

[Nathanael]

The area which the force is applied to is the same between the two systems (the one with two weights and the one with a single weight) therefore the forces will be equal if (and only if) the pressure is equal in the two systems.

Even though one of the systems has twice the weight being applied to it, it is also being applied to (what appears to be) twice the area, and so the pressure inside the two systems is the same.

And so the force must be the same.It can be tricky because of a tendency to look at just the forces involved, but the forces only interact with each other through the fluid (and not directly) so you must think of it in terms of pressure.

 

[BlackSideburns]

ohhh that makes sense. I totally overlooked the fact that there is double the area, thanks. My only concern is how would I show that in terms of formulas/numbers? I would be required to show my work on a test

 

[Nathanael]

ohhh that makes sense. I totally overlooked the fact that there is double the area, thanks. My only concern is how would I show that in terms of formulas/numbers? I would be required to show my work on a test

If I was required to show my work I might put something like this:

A$_{f}$(600g/A$_{w}$) = A$_{f}$(1200g/2A$_{w}$)

Where A$_{f}$ is the area the force is applied to and A$_{w}$ is the area the weights are applied to (and g is the acceleration of gravity)

 

[HalfLostAndSearching]

I would like to provide a response to the content regarding force and pressure on a hydraulic system. In the given diagram, we can see that there are two masses, M1 and M2, acting on a hydraulic system. The hydraulic system consists of a piston with a cross-sectional area A and a force F acting on it.

To determine the forces required to balance the masses, we need to consider the principles of pressure and force. According to the equation P=F/A, the pressure (P) is directly proportional to the force (F) and inversely proportional to the cross-sectional area (A). This means that if we increase the force, the pressure will also increase, and if we decrease the cross-sectional area, the pressure will increase.

Now, let's consider the two FBDs (free body diagrams) for M1 and M2. In both cases, the gravitational force (mg) is acting downwards, while the normal force (N) is acting upwards. In the case of M2, there is also an additional force, F, acting downwards due to the hydraulic system.

When we compare the two FBDs, we can see that for M1, only the normal force is acting upwards, while for M2, the normal force and the force F are acting upwards. This means that the force required to balance M2 (F) will be larger than the force required to balance M1 (N).

In terms of pressure, we can see that the pressure acting on M2 will be larger than the pressure acting on M1. This is because the force F is acting on a smaller cross-sectional area (A) compared to the normal force N. Therefore, the pressure acting on M2 will be larger than the pressure acting on M1.

In summary, the force required to balance M2 will be larger than the force required to balance M1, and the pressure acting on M2 will be larger than the pressure acting on M1. Therefore, the ranking in order from largest to smallest for the forces required to balance the masses (in kgs) would be F, followed by N.