Pulley question - acceleration of an object across a horizontal surface

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The discussion centers on calculating the acceleration of two objects connected by a massless string over a frictionless pulley. The user correctly identifies that the tension in the string is equal for both masses and sets up the equations for net force acting on each mass. The confusion arises from the relationship between tension and weight, particularly for the smaller mass, where the weight exceeds tension, allowing it to accelerate downwards. Ultimately, the user realizes that their approach was correct, but they initially overlooked the fact that the weight of the smaller mass must be greater than the tension for it to accelerate. Understanding this relationship is key to solving the problem accurately.
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Homework Statement


The figure shows two objects of masses m(1) and m(2). The horizontal surface allows for frictionless motion. The string ties to the two objects is massless and passes over a massless pulley that rotates without friction. If m(1)=5.63 kg and m(2)=1.57 kg, what is the magnitude of the acceleration of the sliding object?

http://session.masteringphysics.com/problemAsset/1073792/5/12.P71.jpg

We know that:
- tension m(1) = tension m(2) because they're connected
- mass of m(1), therefore we know the weight because W=mg
- mass of m(2) and we can calculate the weight too
- acceleration is a non-zero value, so Fnet=ma



Homework Equations


Fnet=ma



The Attempt at a Solution


- For the FBD for m(1), the larger object, wouldn't Fnet(x direction) = T= Ma, and Fnet (y direction) = 0 because it's on a flat surface, meaning that the normal and weight cancel out? SO the first equation would be Fnet(x)T=Ma (1)
- For the FBD for m(2), the smaller object, wouldn't Fnet(x direction)=0 since the object is just suspended in the air, and Fnet(y direction)=T-mg... so the 2nd equation would be Fnet(y)T=ma+mg (2)

(I equated Fnet in each axis to mass object * acceleration because they're both moving in that direction)

And since the Tension is the same, you could equate them to each other.. I equated
m(1)a=m(2)a+m(2)g, then isolated for a, but got the wrong answer. What am I doing wrong?!
 
Last edited:
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For your smaller object mg is bigger than T otherwise it wouldn't accelerate.
Fnet=mg-T.You can take it from there.
 
Thanks, that helped a lot. So it turns out that my methods of solving the problem were correct, its just that I didn't realize that W(2) > T... which is what allowed it to accelerate in the direction of W(2). Downwards.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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