Difficulty with this Problem involving a Pully and Cart

[i_love_space_and_eng]

I don’t see any ma’s or I alpha’s. A force balance is an equation. An equation has an = signs.

Alright, so for the weight: T1=mg
For the cart: T2=ma
For the pully, Ia=T1+T2?

 

[Chestermiller]

Alright, so for the weight: T1=mg
For the cart: T2=ma
For the pully, Ia=T1+T2?

Much better. This is actually pretty close. Here are the correct equations:
$$m_1a=m_1g-T_1$$
$$m_2a=T_2$$
$$I\alpha=(T_1-T_2)R$$
where R is the radius of the pulley, $m_1$ is the mass of the crate, $m_2$ is the mass of the cart, I is the moment of inertia of the pulley, $T_1R$ is the moment of the force $T_1$ about the axle of the pulley, and $T_2R$ is the opposing moment of the force $T_2$ about the axle of the pulley.

These equations are what you get if you do analyze the problem thoughtfully, without trying to rush it through. Please, in all the force balances you write in the future, have ma's on the left hand side of the equation and have equal signs.

Do these equations make sense to you? Geometrically, given that the radius of the pulley is R, how is the angular acceleration $\alpha$ related to the acceleration a of the two masses?

 

[haruspex]

Okay...
For T1, it would be clockwise. and for T2, the same.

T2 acts at the top of the pulley and pulls to the left. How is that clockwise?

 

[i_love_space_and_eng]

Much better. This is actually pretty close. Here are the correct equations:
$$m_1a=m_1g-T_1$$
$$m_2a=T_2$$
$$I\alpha=(T_1-T_2)R$$
where R is the radius of the pulley, $m_1$ is the mass of the crate, $m_2$ is the mass of the cart, I is the moment of inertia of the pulley, $T_1R$ is the moment of the force $T_1$ about the axle of the pulley, and $T_2R$ is the opposing moment of the force $T_2$ about the axle of the pulley.

These equations are what you get if you do analyze the problem thoughtfully, without trying to rush it through. Please, in all the force balances you write in the future, have ma's on the left hand side of the equation and have equal signs.

Do these equations make sense to you? Geometrically, given that the radius of the pulley is R, how is the angular acceleration $\alpha$ related to the acceleration a of the two masses?

Hm... Okay, they actually do make a lot of sense! My only question is, for the m_1a=m_1g-T_1 equation, what would a be? Because isn't the acceleration of the weight dependent on gravity?

 

[i_love_space_and_eng]

T2 acts at the top of the pulley and pulls to the left. How is that clockwise?

Sorry, yes, I see how it is counterclockwise!

 

[haruspex]

Hm... Okay, they actually do make a lot of sense! My only question is, for the m_1a=m_1g-T_1 equation, what would a be? Because isn't the acceleration of the weight dependent on gravity?

Yes, that's the m₁g term, but it is partly countered by the tension in the string.

 

[i_love_space_and_eng]

Yes, that's the m₁g term, but it is partly countered by the tension in the string.

Alright... so how would we go on to find the "a" without the T1 and T2?

 

[haruspex]

Alright... so how would we go on to find the "a" without the T1 and T2?

You need one more equation, a simple relationship between a and α.

 

[i_love_space_and_eng]

Okay, so this is what I set up.
I(a/r)= ((m1g-m1a)-m2a)R
Does that look correct? I substituted a/r for α .

 

[Chestermiller]

What is I in terms of the mass M and radius R of the pulley?

 

[i_love_space_and_eng]

What is I in terms of the mass M and radius R of the pulley?

It will be 6.125*10^-5kgm^2

 

[Chestermiller]

It will be 6.125*10^-5kgm^2

Please provide it algebraically.

 

[i_love_space_and_eng]

I=(2/5)mr²

 

[haruspex]

I=(2/5)mr²

So combine the three equations in #32 with those in #39 and #43.

 

[i_love_space_and_eng]

Okay so I plugged it all in and got this:
(2/5)m(r^2)(a/r)=(((m1g)-(m1a))-(m2a))R
I plug in all the correct numbers but the issue is I still get the wrong answer. Did I write my equation correctly? Should I be getting the right answer and I am probably doing a math error?

 

[haruspex]

Okay so I plugged it all in and got this:
(2/5)m(r^2)(a/r)=(((m1g)-(m1a))-(m2a))R
I plug in all the correct numbers but the issue is I still get the wrong answer. Did I write my equation correctly? Should I be getting the right answer and I am probably doing a math error?

Your equation looks right. First simplify it, then post that and your calculation steps.

 

[Chestermiller]

Okay so I plugged it all in and got this:
(2/5)m(r^2)(a/r)=(((m1g)-(m1a))-(m2a))R
I plug in all the correct numbers but the issue is I still get the wrong answer. Did I write my equation correctly? Should I be getting the right answer and I am probably doing a math error?

Those should be capital R's on the left side of the equation, since they are the radius of the pulley. They should also then cancel with the R on the right hand side of the equation. Please provide you algebraic results for a (i.e., in terms of algebraic variables).