Newton's Third Law - Using pulleys and slopes

[tastypotato]

Homework Statement

A 5.9 kg box is on a frictionless 40 degree slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.1 kg weight.

A.
What is the tension in the string if the 5.9 kg box is held in place, so that it cannot move?

B.
If the box is then released, which way will it move on the slope?

C.
What is the tension in the string once the box begins to move?

Homework Equations

F_net=m*a

The Attempt at a Solution

A. The mass of the second box times the gravity will give me the solution to part A -> m₂*(9.81)=20.61N
B. If the box is released then it will slide downwards on the slope because there is more mass in m₁
C. This is where I'm stuck. My professor wrote down this equation for us to follow.

-m₁*g*sin$$\theta$$+T=m₁(-a)
and
m₂*g*T=m₂(a)

but I don't know where to go from here.

 

[tiny-tim]

Hi tastypotato!

(have a theta: θ )

A 5.9 kg box is on a frictionless 40 degree slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.1 kg weight.
…
B.
If the box is then released, which way will it move on the slope?

C.
What is the tension in the string once the box begins to move?
…
B. If the box is released then it will slide downwards on the slope because there is more mass in m₁

Sorry, but that's the wrong reason … "more mass" doesn't clinch it … what matter is the equation your professor wrote down for you …

C. …
-m₁*g*sin$$\theta$$+T=m₁(-a)
and
m₂*g*T=m₂(a)

(and btw, this is https://www.physicsforums.com/library.php?do=view_item&itemid=26", not third)

These two equations are the F_total = ma equations for m₁ and m₂ separately

(you have to do them separately) …

the first one is of the components in the direction of the slope, and the second is for the vertical components …

a is the same in both equations because the string has fixed length, so if m₁ goes distance x up the slope, m₂ goes distance vertically down …

so i] can you prove those two equations are correct?

ii] eliminate T, and find a

 

[nlightner]

I would like to clarify that Newton's Third Law states that for every action, there is an equal and opposite reaction. In this scenario, the action is the force exerted by the hanging weight on the string and the reaction is the tension in the string. Therefore, the tension in the string will be the same for both boxes regardless of their masses.

To solve part C, we can use the equation F=ma. In this case, the force acting on the 5.9 kg box is its weight (m1*g) and the tension in the string (T). This results in the equation m1*g-T=m1*a. Similarly, for the 2.1 kg box, the force acting on it is its weight (m2*g) and the tension in the string (T). This results in the equation m2*g+T=m2*a. Since the acceleration for both boxes will be the same (as they are connected by a string), we can equate the two equations and solve for T. This results in T=(m1-m2)*g/2. Therefore, the tension in the string once the box begins to move is 11.7 N.

It is important to note that in this scenario, the box will not slide down the slope as there is no net force acting on it (since the tension in the string balances out its weight). It will remain in place until the tension in the string is released or changed.