Difficulty with this Problem involving a Pully and Cart

[i_love_space_and_eng]

Homework Statement

A cart with a mass of 1.75 kg is pulled by a string over a pulley attached to a hanging
mass of 400 g.
a. What is the acceleration of the cart if the pulley is considered massless?
b. The pulley has a radius of 3.5 cm and a mass of 100 g. It can be considered a
solid disc. What is the actual acceleration of the cart?

Relevant Equations

Attached

My attempt is also attached. Basically, I think I have (a) down. The only issue is I am getting the wrong answer, and I was hoping one of you could let me know if the answer key is wrong or the way I did it was wrong? The answer is 1.823m/s^2. Also, for the second part, I am really not sure where to go from here. If someone could get any advice that'd be awesome. (Also, the answer to (b) is 1.782m/S^2.

 

[haruspex]

Problem Statement: A cart with a mass of 1.75 kg is pulled by a string over a pulley attached to a hanging
mass of 400 g.
a. What is the acceleration of the cart if the pulley is considered massless?
b. The pulley has a radius of 3.5 cm and a mass of 100 g. It can be considered a
solid disc. What is the actual acceleration of the cart?
Relevant Equations: Attached

My attempt is also attached. Basically, I think I have (a) down. The only issue is I am getting the wrong answer, and I was hoping one of you could let me know if the answer key is wrong or the way I did it was wrong? The answer is 1.823m/s^2. Also, for the second part, I am really not sure where to go from here. If someone could get any advice that'd be awesome. (Also, the answer to (b) is 1.782m/S^2.

Please see post #2 at your prior posting of this problem, https://www.physicsforums.com/threa...-moment-of-inertia-problem-difficulty.972618/.
Also, please do not post algebraic working as an image. That is for diagrams and textbook extracts. Take the trouble to type in your work. That is generally much easier to read and allows those responding to quote specific lines for reference.

 

[i_love_space_and_eng]

Please see post #2 at your prior posting of this problem, https://www.physicsforums.com/threa...-moment-of-inertia-problem-difficulty.972618/.
Also, please do not post algebraic working as an image. That is for diagrams and textbook extracts. Take the trouble to type in your work. That is generally much easier to read and allows those responding to quote specific lines for reference.

Sorry, was told to post again with writing everything. I'll be sure to write out the work next time.

 

[i_love_space_and_eng]

Sorry, was told to post again with writing everything. I'll be sure to write out the work next time.

So, I tried setting it up as ma for cart = ma for the crate, but I go the wrong answer. Could you let me know where I am going wrong with that?

 

[haruspex]

So, I tried setting it up as ma for cart = ma for the crate, but I go the wrong answer. Could you let me know where I am going wrong with that?

As I posted, draw separate FBDs for the hanging block and the cart. Assign an unknown T as the tension. Write out the ΣF=ma equation for each

 

[i_love_space_and_eng]

Alright,
Here i have my fbd1 and fbd2 attached. Any advice on where to continue from here?

 

[Chestermiller]

Alright,
Here i have my fbd1 and fbd2 attached. Any advice on where to continue from here?

$m_2a_2$ is not a force acting on the cart. How are the magnitudes of the accelerations of the cart and the crate related?

 

[i_love_space_and_eng]

$m_2a_2$ is not a force acting on the cart. How are the magnitudes of the accelerations of the cart and the crate related?

Are they related by the force of the string connecting them?

 

[Chestermiller]

Are they related by the force of the string connecting them?

How are they related kinematically, given that the length of the string is constant?

 

[i_love_space_and_eng]

I would say the gravity pulls on the mass, and the mass pulls on the string, which pulls on the cart?

 

[Chestermiller]

I would say the gravity pulls on the mass, and the mass pulls on the string, which pulls on the cart?

So geometrically how are the acceleration magnitudes related? (I don’t want to hear about gravity, masses, or forces)

 

[haruspex]

I would say the gravity pulls on the mass, and the mass pulls on the string, which pulls on the cart?

That does not express the fact that the string stays the same length. What relationship between the two accelerations does express that?
And as Chet notes, your ∑F=ma for the cart is wrong. m₂a₂ is not an applied force.

 

[i_love_space_and_eng]

So geometrically how are the acceleration magnitudes related? (I don’t want to hear about gravity, masses, or forces)

Are the acceleration magnitudes the same? I don't know how not to mention gravity when that is what is pulling on the mass.

 

[i_love_space_and_eng]

That does not express the fact that the string stays the same length. What relationship between the two accelerations does express that?
And as Chet notes, your ∑F=ma for the cart is wrong. m₂a₂ is not an applied force.

Geometrically, they are at 90degree angle? Or atleast the string is.

 

[i_love_space_and_eng]

Are the acceleration magnitudes the same? I don't know how not to mention gravity when that is what is pulling on the mass.

They are at 90degree angles geometrically?

 

[haruspex]

Geometrically, they are at 90degree angle? Or atleast the string is.

That is not a relationship between the magnitudes of the two accelerations and has nothing to do with the total length of string being constant.

 

[Chestermiller]

Are the acceleration magnitudes the same? I don't know how not to mention gravity when that is what is pulling on the mass.

Yes, the acceleration magmoudes are the same? If the crate moves downward 1 cm, how mich does the cart move horizontally?

 

[i_love_space_and_eng]

That is not a relationship between the magnitudes of the two accelerations and has nothing to do with the total length of string being constant.

I actually think I found a and it makes sense... Now here is my work for (b).
Σt=Ia
T-mg=(6.125*10^-5kgm^2)a
T-3.92N= (6.125*10^-5kgm^2)a
Not sure how to find T though... Did I even set that up correctly? Any advice on where to go from here?

 

[i_love_space_and_eng]

Yes, the acceleration magmoudes are the same? If the crate moves downward 1 cm, how mich does the cart move horizontally?

I actually think I found a and it makes sense... Now here is my work for (b).
Σt=Ia
T-mg=(6.125*10^-5kgm^2)a
T-3.92N= (6.125*10^-5kgm^2)a
Not sure how to find T though... Did I even set that up correctly? Any advice on where to go from here?

 

[haruspex]

I actually think I found a and it makes sense... Now here is my work for (b).
Σt=Ia
T-mg=(6.125*10^-5kgm^2)a
T-3.92N= (6.125*10^-5kgm^2)a
Not sure how to find T though... Did I even set that up correctly? Any advice on where to go from here?

As I wrote in the other thread, for part b you need three FBDs; one for the cart, one for the hanging mass and one for the pulley.
The tensions in the two sections of string are now different.
Again, the accelerations are related, but for the pulley it is an angular acceleration. To avoid confusion, please use a different label for that; α is customary.

 

[i_love_space_and_eng]

As I wrote in the other thread, for part b you need three FBDs; one for the cart, one for the hanging mass and one for the pulley.
The tensions in the two sections of string are now different.
Again, the accelerations are related, but for the pulley it is an angular acceleration. To avoid confusion, please use a different label for that; α is customary.

Great. Okay, so here I have drawn out the fbds for all three.
Is this correct?

 

[Chestermiller]

Great. Okay, so here I have drawn out the fbds for all three.
Is this correct?

Good. Now let's see the force and moment balances that go along with the relevant diagrams.

 

[i_love_space_and_eng]

Alright, I think I have that done. Please see below.

 

[jbriggs444]

You realize that you've been asked to type this stuff in, right? "Writing" means "typing". Taking pictures of handwritten text does not count.

 

[Chestermiller]

Alright, I think I have that done. Please see below.

I want to see 3 separate equations: a force balance on the cart, a force balance on the crate, and a moment balance on the pulley. These equations should have net forces or moments set equal to ma or $I\alpha$ (I don't want to see any $\Sigma F's$).

 

[i_love_space_and_eng]

You realize that you've been asked to type this stuff in, right? "Writing" means "typing". Taking pictures of handwritten text does not count.

I took pictures so it could be easily seen next to the figures. I've typing out otherwise when I was first told.

 

[i_love_space_and_eng]

I want to see 3 separate equations: a force balance on the cart, a force balance on the crate, and a moment balance on the pulley. These equations should have net forces or moments set equal to ma or $I\alpha$ (I don't want to see any $\Sigma F's$).

Alright, here is my best attempt. I am trying to do what you have stated, but I am not sure how without equaling it to something. But here it is:
Force balance on cart: T2 only, because no other force is acting on the cart.
For balance on weight: T1-mg
Moment balance on pulley: I'm a little confused by this. Would it be T1+T2?

 

[haruspex]

Alright, here is my best attempt. I am trying to do what you have stated, but I am not sure how without equaling it to something. But here it is:
Force balance on cart: T2 only, because no other force is acting on the cart.
For balance on weight: T1-mg
Moment balance on pulley: I'm a little confused by this. Would it be T1+T2?

For each tension, consider whether the torque it exerts on the pulley is clockwise or anticlockwise.

 

[Chestermiller]

Alright, here is my best attempt. I am trying to do what you have stated, but I am not sure how without equaling it to something. But here it is:
Force balance on cart: T2 only, because no other force is acting on the cart.
For balance on weight: T1-mg
Moment balance on pulley: I'm a little confused by this. Would it be T1+T2?

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

 

[i_love_space_and_eng]

For each tension, consider whether the torque it exerts on the pulley is clockwise or anticlockwise.

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

 

[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 dependant 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 dependant 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).