How Do You Calculate the Acceleration of a Mass in a Pulley System?

[rk2658]

Homework Statement

A mass M1 of 3kG is suspended from one corner by a fixed rope, 1, and from another corner by a rope, 2 which passes over a pulley and is connected to a mass M2 of 2kG, and
suppose that at time t = 0 both masses are at rest and the angles made by the
ropes are each π/4 = 45 degrees. Neglect friction in the pulley and the mass of the rope.
This situation is not stable. The blocks will start to move. Please determine the
acceleration of mass 1.

A picture of this is on this page
http://phys.columbia.edu/~millis/1601/assignments/PHYC1601Fall2011Assignment4.pdf

Homework Equations

F= ma

The Attempt at a Solution

Okay so I was able to break up to the forces for both masses.

For the mass of 3 kgs:
Fy= (T1+T2)sin(45) - 30= 3ay
Fx= (T2-T1)cos(45)= 3ax

For the mass of 2 kgs:
Fy=T1-20= 2a

Also since the accelerations of both masses must be the same I know that:
ay^2 + ax^2 =a ^2

Also I'm supposed to use the fact that since one rope is attached to a wall it doesn't move so that the distance traveled only really happens with pulley.

I don't know what to do from here, any help would be greatly appreciated. Thank you.

 

[NascentOxygen]

Okay so I was able to break up to the forces for both masses.

For the mass of 3 kgs:
Fy= (T1+T2)sin(45) - 30= 3ay
Fx= (T2-T1)cos(45)= 3ax

ax=ay=a.cos(45)
so replace these in the above equations.

For the mass of 2 kgs:
Fy=T1-20= 2a

using this result, substitute for T1 in the above pair of equations

And you are left with 2 equations in 2 unknowns. Solve for a.

 

[rk2658]

why is ax=ay= a cos(45) ?

 

[NascentOxygen]

why is ax=ay= a cos(45) ?

Your question has caused me to look at this more closely. I'm not completely confident that I have it right, even now.

Fixed by a rope on the left, the C of G of M1 is constrained to swing in an arc about that fixed rope's anchor point. With the geometry of the diagram, M1's tangential motion currently is along a 45 deg line.

While I think you had it right when you wrote:

Also since the accelerations of both masses must be the same I know that:
ay^2 + ax^2 =a ^2

it is too general an expression, and turns out to be insufficient [for me] to solve for a. It doesn't include everything we know about ax and ay.

Subtract the two equations to eliminate T2. Then substitute for T1, leaving you with an equation in a, ax, and ay.

Fy= (T1+T2)sin(45) - 30= 3ay
Fx= (T2-T1)cos(45)= 3ax

It looks like it's almost solvable using ay2 + ax2=a2, but is difficult. But if you set horiz and vert components of a to be equal, it's easy.

 

[rwisz]

Dear student,

Thank you for reaching out for help with your homework. Figuring out pulleys can be challenging, but with some guidance, you can solve this problem successfully.

To begin, let's consider the forces acting on each mass. For the mass of 3 kg, as you correctly identified, there are two forces: the tension in rope 1 and the tension in rope 2. These forces act in opposite directions and are balanced by the weight of the mass, mg = 30 N. As you noted, we can break these forces into their x- and y-components. In the x-direction, the forces are balanced, so we have:

(T2-T1)cos(45)= 0

Solving for T2, we get T2=T1. This means that the tensions in both ropes are equal in magnitude.

In the y-direction, the forces are not balanced, so we have:

(T1+T2)sin(45) - 30 = 3ay

Substituting T2=T1, we get:

2T1sin(45) - 30 = 3ay

We also know that the acceleration in the y-direction is equal to the acceleration in the x-direction (since the ropes are connected to the same mass), so we can rewrite this equation as:

2T1sin(45) - 30 = 3ax

Now, let's look at the forces acting on the mass of 2 kg. We have the tension in rope 1 and the weight of the mass, mg = 20 N. Again, we can break these forces into their x- and y-components. In the y-direction, the forces are balanced, so we have:

T1 - 20 = 2a

And in the x-direction, the force is balanced, so we have:

T1cos(45) = 0

Solving for T1, we get T1=0.

Now, we can use this value of T1 in our equation for the y-direction to find the acceleration of the mass of 2 kg:

0 - 20 = 2a

Solving for a, we get a = -10 m/s^2. This negative sign indicates that the mass is accelerating downward.

Finally, we can use this value of a in our equation for the x-direction to find the acceleration of the mass of 3 kg:

2T1sin(45) -