Using acceleration to find tension (concept problem)

[SkyOfMyOwnLight]

Homework Statement

Two boxes of masses M1= 2kg and M2 = 1 kg are pulled up a frictionless ramp by a force P=25N. Calculate the tension in the rope connecting the boxes. Theta is 40 degrees.

Relevant Equations

F = m* a

Hi guys, I feel that this question has a very straightforward answer but I am just not quite grasping it. The first picture is the homework question, the second is the free body diagram I drew.

I know that a = F/m and I know the solution should be:

a = (P - m1*gsinθ - m2*gsinθ) / m1 + m2

What I don't understand is why isn't tension (T) included in the above equation? It's clearly along the x-axis with the others and P is used, so why not tension as well?

The equation that seems logical to me to use is:

a = (T + P - m1*gsingθ - m2*gsingθ ) / m1 + m2

but this is wrong. Can you help me understand why T doesn't belong?

Thank you in advance!

 

[nasu]

The tension acts on both objects not just on one of them. It is an internal force if you consider the system of the two objects.

 

[Chestermiller]

If you we’re doing a force balance on each body individually, the tension T would come into each of the force balances. What would those two force balance equations be?

 

[SammyS]

Problem Statement: Two boxes of masses M1= 2kg and M2 = 1 kg are pulled up a frictionless ramp by a force P=25N. Calculate the tension in the rope connecting the boxes. Theta is 40 degrees.
Relevant Equations: F = m* a

Hi guys, I feel that this question has a very straightforward answer but I am just not quite grasping it. The first picture is the homework question, the second is the free body diagram I drew.

I know that a = F/m and I know the solution should be:

a = (P - m1*gsinθ - m2*gsinθ) / m1 + m2

What I don't understand is why isn't tension (T) included in the above equation? It's clearly along the x-axis with the others and P is used, so why not tension as well?

The equation that seems logical to me to use is:

a = (T + P - m1*gsingθ - m2*gsingθ ) / m1 + m2

but this is wrong. Can you help me understand why T doesn't belong?

Thank you in advance!

Hello, @SkyOfMyOwnLight .



The given solution treats the two boxes along with the connecting rope as one object.

By the way: the sum of the masses, m₁ + m₂, should be enclosed in parentheses if you mean for it all to be in the denominator as in the following.

a = (P - m₁⋅g⋅sinθ - m₂⋅g⋅sinθ) / (m₁ + m₂)

This acceleration can then be used to determine the tension, T .

If you want to use your second image for free body diagrams,

you are missing that the rope also exerts a tension, T, on the box of mass, m₂ . (Yes, then what you have would be two free body diagrams in one image.) You get two equations, one for each box.