Homework Help: How do you find tension and friction on an inclined plane

1. Aug 31, 2015

Leah200

1. The problem statement, all variables and given/known data
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/

2. Relevant equations
Fg = ma

3. 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

2. Aug 31, 2015

Staff: Mentor

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.

3. Aug 31, 2015

Leah200

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.

4. Sep 1, 2015

Staff: Mentor

Fnet is not zero.

5. Sep 2, 2015

Leah200

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.

6. Sep 2, 2015

Staff: Mentor

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.

7. Sep 2, 2015

Hamza Abbasi

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

8. Sep 2, 2015

Leah200

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.

9. Sep 2, 2015

Leah200

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

10. Sep 2, 2015

Staff: Mentor

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

Unrelated issue:
Make sure you use signs consistently.

11. Sep 2, 2015

Leah200

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?

12. Sep 2, 2015

SammyS

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

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

13. Sep 3, 2015

Staff: Mentor

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?

14. Sep 3, 2015

Leah200

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.

15. Sep 3, 2015

Staff: Mentor

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.

16. Sep 3, 2015

SammyS

Staff Emeritus
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 - FT, 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.

17. Sep 3, 2015

Leah200

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

18. Sep 3, 2015

SammyS

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

19. Sep 8, 2015

Leah200

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?

20. Sep 9, 2015

HarperXZC

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

21. Sep 9, 2015

Staff: Mentor

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

22. Sep 9, 2015

SammyS

Staff Emeritus
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?

23. Sep 10, 2015

Leah200

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

24. Sep 10, 2015

SammyS

Staff Emeritus
Yes.

25. Sep 10, 2015

Leah200

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?