Two masses on a slope with a connected tension force

[willis123]

Homework Statement

Consider two blocks hanging on a frictionless incline where theta one is 60 and theta two is 24, what mass does the block on the right need to have in order for the system to stay at rest

Relevant Equations

F=ma and free body with forces

I derived an equation but I have no idea how I can find the mass of the unknown without the normal force in either the x or y direction

 

[Mark44]

Homework Statement: Consider two blocks hanging on a frictionless incline where theta one is 60 and theta two is 24, what mass does the block on the right need to have in order for the system to stay at rest
Relevant Equations: F=ma and free body with forces

View attachment 340226
View attachment 340227
I derived an equation but I have no idea how I can find the mass of the unknown without the normal force in either the x or y direction

Yes, you derived an equation, but 1) it's very difficult to read what you wrote due to the angle you took the picture at, and 2) your equation has almost nothing to do with the problem at hand.
For the 6Kg mass, the force directly down is 6.0*g. Since the angle on that ramp is given, you should be able to calculate the force down the ramp and the force normal to the ramp. You'll need some simple trig to do this, not what it seems you've used (Pythagoras?). Set up a force diagram for the other mass.
For the system to be in equilibrium, which forces need to have equal magnitudes?

Note that g, the acceleration due to gravity, can be cancelled from the equation you get.

 

[Steve4Physics]

Hi @willis123 and welcome to PF. It's worth reading the forum rules: https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

We're assuming there is no friction. Here are a couple of questions for you:

1. Whch is greater - the tension in the left section of the rope or the tension in the right section of the rope? Or are the 2 tensions equal? Or can't you tell until the calculation is done?

2. Do you know how to resolve the weight of, say the mass on the left (call it $m_1$) into a component parallel to the slope (with angle $\theta_1$) and a component perpendicular to the slope? If so, what is the component parallel to the slope in terms of $m_1, g$ and $\theta_1$?

Once you've answered the above questions, you might be able to proceed. If not, post your answers to help us understand where you are at (giving away my age).

Edit - I only just noticed @Mark44's reply. My reply overlaps his.

 

[willis123]

Basically I am going to set the two Tensions equal to eachother because the system is not accelerating then I found the x and y components of both blocks like the normal force in both directions so I can set the y direction stuff like Ny and Ty equal to MG I just don’t know how I can solve my equations without having the normal force component of it in order to take the sin and cosine of it.
Sorry for not following the rules on my previous post hopefully this is better :)

 

[Steve4Physics]

View attachment 340239
Basically I am going to set the two Tensions equal to eachother because the system is not accelerating

Note that the two tensions are equal even if there is acceleration.

(For information, the above statement is true providing:
- the rope has negligible mass;
- the rope can slide over the pulley-wheel with no friction, or
- there is friction between the rope and the pulley-wheel but the pulley-wheel has a negligible moment of inertia (so doesn't 'resist' being rotated).)

then I found the x and y components of both blocks like the normal force in both directions so I can set the y direction stuff like Ny and Ty equal to MG I just don’t know how I can solve my equations without having the normal force component of it in order to take the sin and cosine of it.

You are taking the x-axis to be horizontal and the y-axis to be vertical. But the ‘trick’ is this. For each mass, take:
- the x-axis as parallel to the slope;
- the y-axis as normal to the slope.
Then when you resolve a mass's weight into x and y-components you have this:

(Image from https://www.vedantu.com/question-sets/c1b135ef-7628-468a-a9d7-3c2160aa8c658341706160038218159.png)

By the way, your written working is hard to read/follow. If you can learn to use LaTeX for equations it will help a lot (and you'll get more replies!). The LaTex Guide link is at bottom left of the edit-window.

 

[BvU]

I tried to decipher the scribblings but gave up. But I can read $T_1=T_2$ .
I do notice a lot of $N$ and I wonder why you insist on hanging on to those.

$T_1$ must compensate the component of $m_1g$ along the slope. See Steve's picture. That way you determine $T$ and the same relationship on the other side (now with $T$ known) gives you $m_2$. Done.

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