How do you find tension and friction on an inclined plane

[Leah200]

Homework Statement

How do you find tension and friction on an inclined plane given acceleration, two masses and an angle? It's a pulley system, I suppose.

Here's a diagram I drew:
http://postimg.org/image/m9v9i2qy5/

Homework Equations

Fg = ma

The Attempt at a Solution

I was thinking that I should find the tension for the first mass using Fnet = Fg + FT, then I thought I would find the friction using μ = tan θ, then I would find the second mass' tension through finding the normal force and then using Fnet= Fg - FT - Ff

 

[mfb]

I was thinking that I should find the tension for the first mass using Fnet = Fg + FT

That works.

then I thought I would find the friction using μ = tan θ

Where does that equation come from?

then I would find the second mass' tension through finding the normal force and then using Fnet= Fg - FT - Ff

You can use this equation to calculate the force from friction and then work backwards to get the coefficient of friction.

 

[Leah200]

That works.
Where does that equation come from?
You can use this equation to calculate the force from friction and then work backwards to get the coefficient of friction.

The equation (μ = tan θ) comes from Fnet = mgsinθ - μmgcosθ. Supposedly, I can use it to calculate the coefficient of kinetic friction.

I'll try that out. Thanks!

How would solve it if I had to find the net tension of the two masses? I think the first one would just be Fnet = Fg -/+FT, and solve for FT. Then for the second, would I just find the coefficient of kinetic friction using the angle and tan or would I find the tension using Fnet = Fg -/+FT, and then find the coefficient of kinetic friction using Fnet=Fg - FT-Ff.

 

[mfb]

The equation (μ = tan θ) comes from Fnet = mgsinθ - μmgcosθ

Fnet is not zero.

How would solve it if I had to find the net tension of the two masses?

You found tension already.

 

[Leah200]

Fnet is not zero.
You found tension already.

So, in order to find tension and friction, I just need to use ma = mg + FT for the first mass (the one hanging), then find the second tension by first finding normal force through using FN = mgcosθ, then find the coefficient of friction using μ = tan θ, then use Ff = μFN, and finally use ma = Fg - FT - Ff to find the second tension. Then add the two tensions together?

Sorry about asking so many dumb questions, I just feel lost. I had this problem on a test and lost a ton of points to it, so I wanted to understand it better.

 

[mfb]

There is no second tension, I think the pulley is without friction. The rope has the same tension everywhere.

then find the coefficient of friction using μ = tan θ

That does not work, as Fnet is not zero.

 

[Hamza Abbasi]

Homework Statement

How do you find tension and friction on an inclined plane given acceleration, two masses and an angle? It's a pulley system, I suppose.

Here's a diagram I drew:
http://postimg.org/image/m9v9i2qy5/

Homework Equations

Fg = ma

The Attempt at a Solution

I was thinking that I should find the tension for the first mass using Fnet = Fg + FT, then I thought I would find the friction using μ = tan θ, then I would find the second mass' tension through finding the normal force and then using Fnet= Fg - FT - Ff

Can you please post the complete question with all the numerical values? I also had this type of question in test.

 

[Leah200]

Can you please post the complete question with all the numerical values? I also had this type of question in test.

It went something like this:
Two blocks are connected by a pulley, one block (block 2, mass is 3.8kg) is on an inclined plane of 40 degrees while another block (block 1, mass is 1.8kg) is hanging by the pulley. Block 2 is accelerating up the plane at 1.2m/s^2. Find the tension and friction.

 

[Leah200]

There is no second tension, I think the pulley is without friction. The rope has the same tension everywhere.
That does not work, as Fnet is not zero.

The question asks for friction, unfortunately. It states I need to find tension and then friction.

 

[mfb]

The question asks for friction, unfortunately. It states I need to find tension and then friction.

Yes, friction between the block and the surface, not friction in the pulley.

Unrelated issue:

I was thinking that I should find the tension for the first mass using Fnet = Fg + FT

Make sure you use signs consistently.

 

[Leah200]

Yes, friction between the block and the surface, not friction in the pulley.

Unrelated issue:
Make sure you use signs consistently.

According to my course, when I look for acceleration and tension, I have to use the equations ma = mg + FT for the first mass (the one hanging), then for the second mass (the one on the plane), I have to use FN = mgcos, Ff = miuFN and finally ma = mg - FT - Ff. Instead of finding acceleration, I have to find tension and friction. So, to calculate tension, I just need to use one equation to calculate tension for both masses? Then, to calculate the coefficient of kinetic friction, do I need to then use an equation like this: Ff = FT - mg?

 

[SammyS]

According to my course, when I look for acceleration and tension, I have to use the equations ma = mg + FT for the first mass (the one hanging), ...

It's customary to treat g as a positive quantity, approximately 9.8 m/s², then apply a sign explicitly.

Assuming F_T is upward and gravity acts downward, that results in ma = - mg + F_T.

 

[mfb]

So, to calculate tension, I just need to use one equation to calculate tension for both masses?

That's the point of a frictionless pulley system, tension is the same everywhere.

FN = mgcos, Ff = miuFN

µ or "mu". Apart from that: okay.

ma = mg - FT - Ff

mg would be the force for a completely vertical slope (like for the left mass), here you have a different force. And again, check the signs. Which force is acting in which direction?

 

[Leah200]

That's the point of a frictionless pulley system, tension is the same everywhere.

µ or "mu". Apart from that: okay.mg would be the force for a completely vertical slope (like for the left mass), here you have a different force. And again, check the signs. Which force is acting in which direction?

Oops, I should have written that the hanging mass would be ma = mg - FT, and the acceleration mass would be ma = mg + FT - Ff. I think the force for the hanging mass is negative and the force for the accelerating mass is positive.

 

[mfb]

I think the force for the hanging mass is negative and the force for the accelerating mass is positive.

Start with the actual directions ("up"/"down"), then keep the signs consistent based on that.

and the acceleration mass would be ma = mg + FT - Ff

Why mg? This is wrong. The mass is on an inclined plane.

 

[SammyS]

Oops, I should have written that the hanging mass would be ma = mg - FT, and the acceleration mass would be ma = mg + FT - Ff. I think the force for the hanging mass is negative and the force for the accelerating mass is positive.

The problem states that the block accelerates up the plane. Assuming that's the direction you choose for acceleration, a, to be positive, then it makes sense to consider that for the hanging mass its acceleration is positive it the downward direction. Then your equation for the hanging mass is correct, including the signs. ma = mg - F_T, The tension exerted by the rope on the hanging mass is upward, thus the minus sign.

For the block on the incline, gravity opposes the acceleration. Also, the incline has the implies that the full effect of mg does not belong in the equation for this block.

 

[Leah200]

Start with the actual directions ("up"/"down"), then keep the signs consistent based on that.
Why mg? This is wrong. The mass is on an inclined plane.

Ok, I'll do that. I should have put mgsinθ, I think!

 

[SammyS]

Ok, I'll do that. I should have put mgsinθ, I think!

Right ... and you might want to reconsider the signs in that equation.

 

[Leah200]

Right ... and you might want to reconsider the signs in that equation.

So, for the block on the incline, should I use ma = -mgsinθ + FT to solve for FT. So basically, for the mass hanging use ma = mg - FT, and for the block on the inclined plane, use ma = -mgsinθ + FT? Should I add the two tensions together? If I look for the coefficient of kinetic friction, should I use μ = tan θ?

 

[HarperXZC]

How to solve this problem, if I had to find the two objects net tension?

 

[mfb]

@HarperXZC: If you allow for friction both on the inclined plane and on the pulley, the system is underdetermined - you need more information to solve it then.

If I look for the coefficient of kinetic friction, should I use μ = tan θ?

No. I don't know why you keep repeating this wrong formula.
There are no tensions to add.

 

[SammyS]

So, for the block on the incline, should I use ma = -mgsinθ + FT to solve for FT. So basically, for the mass hanging use ma = mg - FT, and for the block on the inclined plane, use ma = -mgsinθ + FT?

That looks good. The signs are now consistent. However, you are missing a term for friction in the second equation.

Should I add the two tensions together?

No. Just solve two equations in two unknowns.

If I look for the coefficient of kinetic friction, should I use μ = tan θ?

No.

Why would you assume that?

 

[Leah200]

That looks good. The signs are now consistent. However, you are missing a term for friction in the second equation.

No. Just solve two equations in two unknowns.

No.

Why would you assume that?

So for the second mass, it should be ma = -mgsinθ + FT - Ff, solve for FT, then should I solve for Ff?

 

[SammyS]

So for the second mass, it should be ma = -mgsinθ + FT - Ff, solve for FT, then should I solve for Ff?

Yes.

 

[Leah200]

Yes.

Let's say that the angle is 30 degrees, mass is 3.2kg, acceleration is 5.4m/s^2 for the second mass.

So it's (3.2kg)(5.4m/s^2) = - (3.2kg)(9.8m/s^2)sin30 + FT - Ff, then 17.28kg/s^2 = -15.68kgm/s^2 + FT - Ff, then can I add them like this 32.96kgm/s^2 = FT - Ff? After that, can I then solve for it like this 32.96kgm/s^2/FT = -Ff and then insert the first mass' value of FT into this equation? After that, solve for μ using Ff = μFN?

 

[SammyS]

... , then can I add them like this 32.96kgm/s^2 = FT - Ff? After that, can I then solve for it like this 32.96kgm/s^2/FT = -Ff
...

That's not a valid algebra step.

 

[Leah200]

That's not a valid algebra step.

Okay. Should I find friction before I find tension? The question asked me to find tension and then solve for the coefficient of kinetic friction.

 

[SammyS]

Okay. Should I find friction before I find tension? The question asked me to find tension and then solve for the coefficient of kinetic friction.

It seems reasonable to find tension first, especially with this particular set of equations.

The point I was making was that you divided by F_T, which really doesn't do what you want.

 

[HarperXZC]

How to solve this problem, if I had to find the two objects net tension?
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[haruspex]

That's the point of a frictionless pulley system, tension is the same everywhere.

Hmmm... not exactly.
When a pulley is described as frictionless, that should be taken only as referring to the bearing. Conversely, one should not expect any slippage between pulley and rope. Since the system is accelerating, for the tension to be the same both sides we also need to assume the pulley is massless.

 

[mfb]

A fully frictionless pulley system would slip between pulley and rope, making the pulley mass irrelevant.
Anyway, no information about the pulley system is given, so frictionless and massless is the most reasonable assumption.

 

[haruspex]

A fully frictionless pulley system would slip between pulley and rope, making the pulley mass irrelevant.

Yes, but that is never what is meant in a problem statement by a 'frictionless pulley'. If it were frictionless in that sense there would be no need for it to be a pulley - it could just be a frictionless shoulder.

Anyway, no information about the pulley system is given, so frictionless and massless is the most reasonable assumption.

Sure, but Leah (or other readers) may in future encounter problems where the pulley has inertia, so I felt it was important to get the statement right. (And problems where the pulley has relevant inertia may also describe the pulley as frictionless, without necessarily clarifying that this only refers to the axle.)