#### rk2658

1. 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.

http://phys.columbia.edu/~millis/1601/assignments/PHYC1601Fall2011Assignment4.pdf

2. Homework Equations
F= ma

3. 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 travelled only really happens with pulley.

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

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#### NascentOxygen

Mentor
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

Mentor
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.

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