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#### deleted647690

In EK1001, I was trying to do a problem that showed a basic inclined plane with a pulley on a box with some tension T. "The plane above is inclined at a 30 degree angle. The coefficient of kinetic friction between the mass and the plane is 0.1. The mass is 100 kg. What is the minimum tension required in the rope so that the mass will accelerate down the plane at 2 m/s^2?"

I used a = Fnet / m
The force of friction and tension and both acting up, while the mgsintheta component is acting downwards.

So 2 m/s^2 = Force of friction + Tension - mgsin theta / m

Force of friction = u * mg cos theta
Force of friction = 86

2 m/s^2 = 86 + T + -500 / 100

T = 614

This was not correct. The correct answer was 213 N. In the key, they set up mgsintheta = u mgcostheata + T + ma. What is this "ma" term? I am confused about what this additional force is that they are adding into the equation. Why does a = fnet/m not work with the way I had it set up?

Thank you

I guess my real question is about how to set up the net forces correctly. I keep messing those up.

I'm not really sure what the "ma" term means in this question. If the box is accelerating downwards, shouldn't "ma" be added to the downward force side?

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#### rhinzaz

Hello,

Great question;

The term 'ma' used in the explanation is essentially the Fnet (the force of acceleration) which is (100kg)(2m/s^2). So it's not an 'additional force' they are adding.

The issue with your equation is not the way you set it up (in terms of directionality); however, your equation is half-simplified per se; you can see that by noticing how you're relating 2m/s^2 = Force of friction + Tension - mgsin theta/m. Acceleration does not equal Force directly, and your units do not match.

If you had done this, rather; ma = Force of friction + Tension - mgsign theta/m, you should see that you might arrive at the correct answer.

Just remember that your units have to match whenever you're equating terms

#### rhinzaz

Hello,

Great question;

The term 'ma' used in the explanation is essentially the Fnet (the force of acceleration) which is (100kg)(2m/s^2). So it's not an 'additional force' they are adding.

The issue with your equation is not the way you set it up (in terms of directionality); however, your equation is half-simplified per se; you can see that by noticing how you're relating 2m/s^2 = Force of friction + Tension - mgsin theta/m. Acceleration does not equal Force directly, and your units do not match.

If you had done this, rather; ma = Force of friction + Tension - mgsign theta/m, you should see that you might arrive at the correct answer.

Just remember that your units have to match whenever you're equating terms
Actually I was wrong; your directionality does appear to have a small issue as well.

The way you wrote it; 2m/s^2 = Force of friction + Tension - mgsin theta/m,

First, we need to fix it by replacing a with Fnet (ma); as discussed above, so it becomes;

Fnet = 100kg x 2m/s^2 = Force of friction + Tension - mgsin theta/m

The problem now is, based on the way you wrote your equation, it is implying that the Fnet is actually in the same direction as the Tension force and force of friction. Another way to think about it is saying "the force of friction plus tension subtract force of gravity results in a net force (the leftover) that is pointing the same way as Ff/Tension.

D

#### deleted647690

Hello,

Great question;

The term 'ma' used in the explanation is essentially the Fnet (the force of acceleration) which is (100kg)(2m/s^2). So it's not an 'additional force' they are adding.

The issue with your equation is not the way you set it up (in terms of directionality); however, your equation is half-simplified per se; you can see that by noticing how you're relating 2m/s^2 = Force of friction + Tension - mgsin theta/m. Acceleration does not equal Force directly, and your units do not match.

If you had done this, rather; ma = Force of friction + Tension - mgsign theta/m, you should see that you might arrive at the correct answer.

Just remember that your units have to match whenever you're equating terms

edit

D

#### deleted647690

Hello,

Great question;

The term 'ma' used in the explanation is essentially the Fnet (the force of acceleration) which is (100kg)(2m/s^2). So it's not an 'additional force' they are adding.

The issue with your equation is not the way you set it up (in terms of directionality); however, your equation is half-simplified per se; you can see that by noticing how you're relating 2m/s^2 = Force of friction + Tension - mgsin theta/m. Acceleration does not equal Force directly, and your units do not match.

If you had done this, rather; ma = Force of friction + Tension - mgsign theta/m, you should see that you might arrive at the correct answer.

Just remember that your units have to match whenever you're equating terms

So it was incorrect to rearrange f = ma to a=fnet/m??

D

#### deleted647690

Actually I was wrong; your directionality does appear to have a small issue as well.

The way you wrote it; 2m/s^2 = Force of friction + Tension - mgsin theta/m,

First, we need to fix it by replacing a with Fnet (ma); as discussed above, so it becomes;

Fnet = 100kg x 2m/s^2 = Force of friction + Tension - mgsin theta/m

The problem now is, based on the way you wrote your equation, it is implying that the Fnet is actually in the same direction as the Tension force and force of friction. Another way to think about it is saying "the force of friction plus tension subtract force of gravity results in a net force (the leftover) that is pointing the same way as Ff/Tension.
If you wrote it as ma = fnet/m, wouldn't the m's cancel and leave you with a = fnet

I think the units for a = fnet/m work out correctly
m/s^2 = m * m/s^2 / m

#### rhinzaz

Hi Acetylmandarin,

Sorry; I was a bit confused when looking at your equation. Based on your comment, I suppose in your equation:

2 m/s^2 = Force of friction + Tension - mgsin theta / m

your numerator is "Force of friction + Tension - mgsin theta" and denominator is m. (I would also suggest to bracket the numerator for clarification; this is where BEDMAS become helpful!)

If that's the case then yes, your units work fine. Where I got confused was your equation I thought you had written it as

2 m/s^2 = (Force of friction) + (Tension) - (mgsin theta / m)

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#### deleted647690

sorry, I meant to edit my original post and then accidentally made a new post

#### rhinzaz

To get to the root of your question; however, requires answering "I'm not really sure what the "ma" term means in this question. If the box is accelerating downwards, shouldn't "ma" be added to the downward force side?"

Best thing to do is start a free body diagram.

Pointing in the downward direction is Fg. Pointing in the upward direction is T and Ff (friction force).

the sum of the forces result in a net force Fnet, so Ff + T + Fg = Fnet

We can arbritarily assign downward direction as negative, and upward direction as positive, so the equation now becomes

Ff + T + (-Fg) = -(fnet)

T = -Ff + Fg - fnet

T = -0.1mgcos30 + mgsin30 - ma
-plugging in equations give

T = -87N + 500N - 200N

T = 213N

With questions like these; it's always best to have a systematic approach, until it becomes second nature to you that you know how to balance the forces mentally.

D

#### deleted647690

To get to the root of your question; however, requires answering "I'm not really sure what the "ma" term means in this question. If the box is accelerating downwards, shouldn't "ma" be added to the downward force side?"

Best thing to do is start a free body diagram.

Pointing in the downward direction is Fg. Pointing in the upward direction is T and Ff (friction force).

the sum of the forces result in a net force Fnet, so Ff + T + Fg = Fnet

We can arbritarily assign downward direction as negative, and upward direction as positive, so the equation now becomes

Ff + T + (-Fg) = -(fnet)

T = -Ff + Fg - fnet

T = -0.1mgcos30 + mgsin30 - ma
-plugging in equations give

T = -87N + 500N - 200N

T = 213N

With questions like these; it's always best to have a systematic approach, until it becomes second nature to you that you know how to balance the forces mentally.
Thank you.
So I would have gotten the correct answer with the original way I posted had I set the acceleration to be negative instead of positive them.
Thank you for showing me this.

So the way you wrote it was to say that ma = the sum of the forces, and I just did the a = the sum of the forces divided by m, so same thing then.

Thank you. What is BEMDAS?
I always used PEMDAS. Please Excuse My Dear Aunt Sally

D

#### deleted647690

Thank you.
So I would have gotten the correct answer with the original way I posted had I set the acceleration to be negative instead of positive them.
Thank you for showing me this.

So the way you wrote it was to say that ma = the sum of the forces, and I just did the a = the sum of the forces divided by m, so same thing then.

Thank you. What is BEMDAS?
I always used PEMDAS. Please Excuse My Dear Aunt Sally

Also, sorry for the unclear equation

#### rhinzaz

Thank you.
So I would have gotten the correct answer with the original way I posted had I set the acceleration to be negative instead of positive them.
Thank you for showing me this.

So the way you wrote it was to say that ma = the sum of the forces, and I just did the a = the sum of the forces divided by m, so same thing then.

Thank you. What is BEMDAS?
I always used PEMDAS. Please Excuse My Dear Aunt Sally

Lol, BEDMAS and PEDMAS are the same thing; both are rules following the order of operation in arithmetic. B=bracket and P=parenthesis, same thing!