How to Find Equilibrium in a Pulley System with Multiple Forces

[-EquinoX-]

Homework Statement

I did an experiment on lab last week, and not really sure if this is right, but please help
me:

There are 3 forces being performed on a pulley framework. The forces are F1 = 50
which is 160 degrees of the x-axis and F2 = 100 which is 60 degrees of the x-axis.
F3 is just a mass hanging on the system with a direction of -y (270 degrees), and the mass is 125g.

I need to show that adding those three Forces will give me a value close to 0. Which means
the three are in equilibrium/balanced.

Homework Equations

The Attempt at a Solution

My attempt here is to find the forces on F1 first, which is:

F1 = 50 * (cos 160) i + 50 * sin (160) j = -46.98 i + 17.10 j
F2 = 100 * (cos 60) i + 100 * sin (60) j = 50 i + 86.60 j
F3 = .125kg * 9.8 m/s^2 = -1.225 j

F1 + F2 + F3 = 3.02 i + 102.475 j

Summing all those forces/vectors I should get a smaller value close to 0. However
what I got is different.

The 102.475 here is not right, the forces reacting on the y direction should be smaller
than 86.60.

Is my experiment a failure? Please help me...

 

[alphysicist]

Hi -EquinoX-,

If the three forces are supposed to approximately cancel (and if I'm understanding the setup correctly) my guess would be that the measurements of 50 and 100 for F1 and F2 are not in Newtons.

 

[-EquinoX-]

I am using a spring balance to read the forces... it said that it should be in equlibrium, but does that mean that the results of those 3 must be equal to 0 when in equilibrium?

 

[-EquinoX-]

I am kind of confused what it means to be in an equilibrium? As you can see from my computation above, the forces on the x-axes is smaller than F1 and F2, which for me it seems to be right. However it doesn't hold true for the y-axis.

 

[tiny-tim]

There are 3 forces being performed on a pulley framework. The forces are F1 = 50
which is 160 degrees of the x-axis and F2 = 100 which is 60 degrees of the x-axis.
F3 is just a mass hanging on the system with a direction of -y (270 degrees), and the mass is 125g.
…
F1 = 50 * (cos 160) i + 50 * sin (160) j = -46.98 i + 17.10 j
F2 = 100 * (cos 60) i + 100 * sin (60) j = 50 i + 86.60 j
F3 = .125kg * 9.8 m/s^2 = -1.225 j

F1 + F2 + F3 = 3.02 i + 102.475 j.

If the three forces are supposed to approximately cancel (and if I'm understanding the setup correctly) my guess would be that the measurements of 50 and 100 for F1 and F2 are not in Newtons.

I am using a spring balance to read the forces... it said that it should be in equlibrium, but does that mean that the results of those 3 must be equal to 0 when in equilibrium?

Hi -EquinoX-!

Yes … equilibrium means zero acceleration, so the net force must be zero.
​

Does a spring balance mean that if you hang a 100g weight from it, it reads 100?

If so, you don't multiply by g, do you?

(btw, how accurate was this experiment … your figures seem to show ±10g, which is highly inaccurate)

 

[-EquinoX-]

Well, the other 2 forces, which is 50N and 100N are measured with the spring balance, however the other one is a mass. So I should multiply that with 9.8.

 

[alphysicist]

Well, the other 2 forces, which is 50N and 100N are measured with the spring balance, however the other one is a mass. So I should multiply that with 9.8.

I was just asking if you were certain that the measurements of 50 and 100 were in Newtons. (If, for example, they were centi-Newtons the forces would approximately cancel.)

So think back to the F2. You say F2 is 100N. That's over 22 pounds--around the weight of three gallons of milk. Is that how much weight you had to create F2?

 

[-EquinoX-]

say if they were mili Newtons, then does this hold?

 

[helpneeded02]

I am doing the same experiment. Can you please explain to me how to you solved your experiment (mathematically)? I don't understand how to add the forces if there are completely different directions, as well as how to know what the sum should be smaller than. Please help me!