How Do Accelerations and Tensions Relate in a Dual Pulley System?

[AdkinsJr]

Homework Statement

An object of mass m_1 hangs from a string that passes over a very light fixed pulley P_1. the string connects to a second very light pulley P_2. A second string pases around this pulley with one end attached to a wall and the other to an object of mass m_2 on a frictionless horizontal table. (a) if a_1 and a_2 are the accelerations, then what is the relationship between these accelerations? find expressions for (b) the tenstions in the strings and (c) the accelerations a_1 and a_2 in terms of the masses m_1 and m_2, and g

Homework Equations

Newton's Laws

The Attempt at a Solution

I can't get much further than writing the laws for the masses:

$$m_1a_1=T_1-m_1g$$

$$m_2a_2=T_2$$

My difficulty with physics problems is figuring out how to mathematically express what is happening. My understanding of this system is that T_1 is pulling on the pully P_2 causing it to accelerate. It is therefore putting tension on the wall as well as the mass m_2, causing the mass to accelerate. I'm assuming that the tension T_2 on the wall is the same as the tension on m_2.

I drew a free body diagram for the pulley, which suggests:

$$m_pa_p=T_1-2T_2$$

If I say that m_p is a "really light mass" then setting it to zero then would suggests $$T_1=2T_2$$, but I'm not comfortable with this and I doubt that I'm taking the correct path here.

I'm sure I need to understand the relationship between a_1 and a_2, which I don't.

 

[soothsayer]

I'm having a hard time visualizing this diagram, the description is a bit ambiguous, I feel. Is there any way you could supply a drawing of some sort?

 

[AdkinsJr]

I'm having a hard time visualizing this diagram, the description is a bit ambiguous, I feel. Is there any way you could supply a drawing of some sort?

Yes, lol. I had a diagram, but I forgot to attach it.

 

[Sourabh N]

For (a), consider mass m_1 moves a distance x and see how much m_2 moves.
For (b) and (c), draw free body diagrams for both masses and use the fact that second pulley is massless (this means net force on second pulley is zero )

 

[AdkinsJr]

For (a), consider mass m_1 moves a distance x and see how much m_2 moves.

I had that in mind, but I'm not sure how to determine that. My intuition says it's half the distance, but I can't show that.

For (b) and (c), draw free body diagrams for both masses and use the fact that second pulley is massless (this means net force on second pulley is zero )

Ok, I've done this and I was in fact taking that approach. For the pulley, my second law equation was:

$$m_pa_p=T_1-2T_2$$
$$0=T_1-2T_2$$
$$T_1=2T_2$$

 

[Sourabh N]

Your intuition is right. When m1 moves a distance x, the second pulley has to move a distance x (so that the string remains taut). Now, m2, say, moves a distance y. y has to be 2*x so that this string remains taut.
Slightly tricky at first, but make sure you understand this.

Once you have the relation between x and y, just differentiate twice to get relation between a1 and a2.

 

[soothsayer]

yeah, that looks right, now to find the acceleration on the two masses, set up Newton's 2nd law, F=ma.

 

[AdkinsJr]

Your intuition is right. When m1 moves a distance x, the second pulley has to move a distance x (so that the string remains taut). Now, m2, say, moves a distance y. y has to be x/2 so that this string remains taut.
Slightly tricky at first, but make sure you understand this.

Once you have the relation between x and y, just differentiate twice to get relation between a1 and a2.

Ok, so the distance traveled by m1 is:

$$x_{m_1}=2x_{m_2}$$

Taking two time derivatives:

$$2a_1=a_2$$

So my system of equations is just:

$$m_1a_1=T_1-m_1g$$
$$2m_2a_1=\frac{1}{2}t_1$$

 

[Sourabh N]

Ok, so the distance traveled by m1 is:

$$x_{m_1}=2x_{m_2}$$

Taking two time derivatives:

$$2a_1=a_2$$

So my system of equations is just:

$$m_1a_1=T_1-m_1g$$
$$2m_2a_1=\frac{1}{2}t_1$$

First equation is right. Second equation isn't.

 

[AdkinsJr]

First equation is right. Second equation isn't.

Oh yeah, $$a_1=2a_2$$, I dont' know why I moved the two to the left side -_-

So the system is:

$$m_1a_1=T_1-m_1g$$

$$m_2\left(\frac{1}{2}a_1\right)=\frac{1}{2}T_1$$

 

[Sourabh N]

Good. Now you have two linear equations in two variables. Solve it.

 

[AdkinsJr]

Good. Now you have two linear equations in two variables. Solve it.

Ok, thanks for your help. I already took mechanics; I've been doing a little review to keep my problem solving skills sharp.