- #1

Tlurk

- 5

- 2

- Homework Statement
- I have a system with two pulleys and three masses where we have two outer masses on each end of the connecting rope and one middle mass at the centre of the rope. The length of the rope, the space between the pulleys and the pulley size are chosen arbitrarily. At first, all masses are equal, but we're supposed to change the middle mass later.

a) Can the system be at rest?

b) If so, what is the position of the three masses?

c) What happens if we change the middle mass?

- Relevant Equations
- Newton's laws.

ΣF = 0

Basic trigonometry.

I used m

System:

FBD:

Note: I believe I have solved a) correctly and am more confused about b).

a) I started with drawing the FBD. Knowing that the net sum has to be zero for the system to be at rest I used the left mass to find the tension on the rope: S

I know that S

It looks like the system can be at rest, I just have to find the angle on the ropes on the middle mass.

I know that sin(θ) = S

θ = arcsin(S

So the system is at rest when the angle between the mass and each rope is 30 degrees.b) I used the radius of pulley r=0,20m, the rope length s=3,0m and the length between pulleys (centre to centre) l=0,80m. Again, the system is symmetric so I will look at the left side. I will also use the connection points between the rope and the masses as their position. A simple diagram:

I can find the length h from a straight line between the pulley tops to the middle mass by doing the following:

tan(30deg) = h / 0,40m

h = 0,40m * tan(30deg) = 0,23. I can then find the rope length to the top of the pulley from the middle mass by:

sin(30deg) = h / c

c =h / sin(30deg) = 0,40m * tan(30deg) / sin(30deg) = 0,462m

Also, the rope length around the pulley is about a quarter circle:

The length from the leftmost point of the pulley and down to the left mass is 1,5m - 0,462m -0,314m = 0,724m ≈ 0,72m.

If we now make the coordinate system have its x-axis along the top of the pulleys, and the y-axis through the centre of the first pulley, the masses would have the following positions:

Left mass: (-0,20, -0,20-0,72) = (-0,20, -0,92)

Middle mass: (0,40, -0,23)

Right mass: (1,0, -0,92)

Is this correct? I feel like I'm missing a function to figure out how much rope goes around the pulley as I believe this will be more than a quarter circle if the middle mass weighs a lot more than the outer masses. But I'm not sure how to get started here.

As for c, I want to make sure I got b correct first, and I think it'll solve itself without being too difficult as the angle the system is at rest is only dependent on the masses and can also be found with arcsin(m

_{1}=m_{2}=15,0kgSystem:

FBD:

Note: I believe I have solved a) correctly and am more confused about b).

a) I started with drawing the FBD. Knowing that the net sum has to be zero for the system to be at rest I used the left mass to find the tension on the rope: S

_{1}= G_{1}= 15kg*9,81m/s^{2}= 147,15 N. Now that I know the tension I moved over to the middle mass. I know G_{2}= 147,15 N and that |S_{1}|=|S_{2}|=|S_{3}|=|S_{4}|. As the system is symmetric on each side I didn't do anything with the x-component of S_{2}and S_{3}on the middle mass, as they cancel out.I know that S

_{2y}+ S_{3y}= G_{2}or rather 2 * S_{2y}= G_{2}to simplify as the system is symmetric. Therefore S_{2y}= G_{2}/2 = 147,15N / 2 = 73,575 N.It looks like the system can be at rest, I just have to find the angle on the ropes on the middle mass.

I know that sin(θ) = S

_{2y}/ S_{2}.θ = arcsin(S

_{2y}/ S_{2}) = arcsin(73,575N / 147,15N) = 30 degrees.So the system is at rest when the angle between the mass and each rope is 30 degrees.b) I used the radius of pulley r=0,20m, the rope length s=3,0m and the length between pulleys (centre to centre) l=0,80m. Again, the system is symmetric so I will look at the left side. I will also use the connection points between the rope and the masses as their position. A simple diagram:

I can find the length h from a straight line between the pulley tops to the middle mass by doing the following:

tan(30deg) = h / 0,40m

h = 0,40m * tan(30deg) = 0,23. I can then find the rope length to the top of the pulley from the middle mass by:

sin(30deg) = h / c

c =h / sin(30deg) = 0,40m * tan(30deg) / sin(30deg) = 0,462m

Also, the rope length around the pulley is about a quarter circle:

*0,40m***π / 4 =**π / 10 = 0,314m.*The length from the leftmost point of the pulley and down to the left mass is 1,5m - 0,462m -0,314m = 0,724m ≈ 0,72m.

If we now make the coordinate system have its x-axis along the top of the pulleys, and the y-axis through the centre of the first pulley, the masses would have the following positions:

Left mass: (-0,20, -0,20-0,72) = (-0,20, -0,92)

Middle mass: (0,40, -0,23)

Right mass: (1,0, -0,92)

Is this correct? I feel like I'm missing a function to figure out how much rope goes around the pulley as I believe this will be more than a quarter circle if the middle mass weighs a lot more than the outer masses. But I'm not sure how to get started here.

As for c, I want to make sure I got b correct first, and I think it'll solve itself without being too difficult as the angle the system is at rest is only dependent on the masses and can also be found with arcsin(m

_{2}/ 2m_{1}). Then I would just follow the process in b).