Solve Force and Tension for Masses in Inclined Plane System | Newton's Laws

[jedjj]

Homework Statement

The figure (attached (5-56))shows a box of mass m2 = 2.60 kg on a frictionless plane inclined at angle θ = 33°. It is connected by a cord of negligible mass to a box of mass m1 = 4.00 kg on a horizontal frictionless surface. The pulley is frictionless and massless.

(a) If the magnitude of horizontal force is 2.3 N, what is the tension in the connecting cord?

(b) What is the largest value the magnitude of may have without the cord becoming slack?

Homework Equations

Newton's Laws

The Attempt at a Solution

Unfortunately this is the part of physics where I became lost in high school and I am having a difficult time putting everything together.

So the first thing that I did was set up axes for each. I set m1 as a normal x and y coordinate system, and m2 with a tilted x-y coordinate system (attached FBD).
So first I attempted to find w2 because that should help me find tension.

\sum F_x=m_2*a_x
T-w_{x2}=m_2*a_x \\w in the x direction of mass 2
w_{x2}=-2.6*sin(33)
I should use this to solve for a_x?

moving on to m1
\sum F_x=m_1*a_x
T+F=m_1*a_x
Do I plug in for a_x because if I do, then I have a T on both sides, in a place that wouldn't be easy to solve?

And even where I am at I believe I have made a huge mistake. I am having problems understanding this. Please help me.

 

[Doc Al]

\sum F_x=m_2*a_x
T-w_{x2}=m_2*a_x \\w in the x direction of mass 2
w_{x2}=-2.6*sin(33)

Careful with signs. Which way are the masses accelerating?

I should use this to solve for a_x?

You need to use the equations for both masses to solve for the acceleration.

moving on to m1
\sum F_x=m_1*a_x
T+F=m_1*a_x

Good.

Do I plug in for a_x because if I do, then I have a T on both sides, in a place that wouldn't be easy to solve?

Once you correct the sign error and rearrange the first equation, it may be easier to see how to eliminate T.

 

[jedjj]

Thank you for the quick response.
I'm not sure why I put that negative sign there, but even after fixing that I'm still stuck.
w_{x2}=2.6*sin(33)

and
w_{x2}-T=m_2*a_x
a_x=\frac{w_{x2}-T}{m_2}

but after plugging in
T=m_1\frac{w_{x2}-T}{m_2}-F
making it difficult to solve for T. Am I not understanding something?
[edits for minor mistakes]

 

[Doc Al]

Thank you for the quick response.
I'm not sure why I put that negative sign there, but even after fixing that I'm still stuck.
w_{x2}=2.6*sin(33)

and
w_{x2}-T=m_2*a_x

Stop right there! Now combine this equation with the equation for m1 to eliminate T. (Write them one after the other and compare them.)

 

[jedjj]

Thank you so much for helping me. I feel like I am close, but I am getting a negative tension (and the magnitude of it is not correct).

w_{x2}-m_2*a_x+F=m_1*a_x
a_x=\frac {w_{x2}+F}{m_1+m_2}

T=w_{x2}-m_2*a_x
but I'm getting -.0479
why?

 

[Doc Al]

I feel like I am close, but I am getting a negative tension (and the magnitude of it is not correct).

w_{x2}-m_2*a_x+F=m_1*a_x
a_x=\frac {w_{x2}+F}{m_1+m_2}

This is correct. What did you get for the acceleration?

T=w_{x2}-m_2*a_x

This is also correct. So if you plug your answer for acceleration into this expression, you should be good to go.

The equations are fine. Just do the arithmetic over. (If it's not an arithmetic error, then you are inputing a wrong value somewhere.)

 

[jedjj]

a_x = (2.6*sin(33)+2.3) / (4+2.6)
a_x = 0.563

T = 2.6*sin(33) - (2.6(.563))
T = -.0477

[physics makes me feel ignorant :( ]

 

[Doc Al]

a_x = (2.6*sin(33)+2.3) / (4+2.6)
a_x = 0.563

You left something out: That should be mg sin(33), not m sin(33).

[physics makes me feel ignorant :( ]

Me too!

 

[jedjj]

Thank you so much!
I think I need to do some practice problems similar to this, because I am having a very hard time with this. (I'm sure I can find plenty on this board :) ).