Two blocks connected by string passing over a pulley on an incline

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Two blocks connected by a string over a pulley are analyzed, with one block on a frictionless incline accelerating at 1.97 m/s². The tension in the string segments was calculated, yielding T2 as 152.6 N and T1 as 117.1 N. The moment of inertia of the pulley was initially calculated as 4.05 kg*m² but later corrected to 0.912 kg*m² after reevaluating the equations. The discussion emphasized the importance of using the correct relationships between tension, acceleration, and moment of inertia. Overall, the calculations highlight the complexities involved in solving problems related to pulleys and inclined planes.
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Homework Statement



Two blocks, as shown below, are connected by a string of negligible mass passing over a pulley of radius 0.225 m and moment of inertia I. The block on the frictionless incline is moving up with a constant acceleration of 1.97 m/s2 on an angle of 35.1 degrees.

m1 = 15.4 kg
m2 = 19.5 kg

A set up of the blocks and incline can be seen below, but the values above are used in all calculations
http://www.webassign.net/pse/p10-71alt.gif



(a) Determine the tensions in the two parts of the string.
T1:
T2:
(b) Find the moment of inertia of the pulley.

Homework Equations





The Attempt at a Solution



K, so I solved for T2 but I am unable to solve T1.

For T2, I set m2g-T2 = m2a
(19.5)(9.8) - T2 = (19.5)(1.97)
T2 = 152.6 N


For T1, I found the x component of acceleration to be 1.97cos35.1 = 1.612 m/s^2
From there I though T1 would be mgsin35.1 but that was wrong. I know I need the acceleration to find T1 (at least I think so) because the question specifically stated the acceleration so it must be used to equate T1 with ma or something like that. Please help!

As for B, I know the moment of inertia of a pulley to be 1/2Mr^2.
Rotational Kinetic energy = 1/2Iw^2 ( I don;t know if this is needed to solve the question)

Please, some guidance for both of the questions would be greatly appreciated!
 
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The values given in the diagram and the problem are different. Why?

T1 - m1g*sinθ = m1a.

And (T2 - T1)*r = I*α where α is the angular acceleration which is equal to a/r.
 
Last edited:
The diagram I found off the internet has different values because it is not from my actual question. I posted it as a pictorial representation of what my question looks like.

Thank you for T1, it worked :)!

Now to try the second part :)
 
Ok, so I tried the second part

T2 - T1 = Ialpha
152.6N - 117.1 N = I (a/r)
35.5 N = I (1.97/0.225)
4.05 kg*m^2 = I

The answer came out to be wrong :( I don't see why this answer is wrong, it makes sense to me. Any ideas?
 
Sorry.

T2 - T1 = I*α/r. = I*a/r^2
 
T2 - T1 = I*α/r. = I*a/r^2

T2 = 152.6N
T1 = 117.1 N

a = 1.97 m/s^2
r = 0.225m

T2 - T1 = I*a/r^2
152.6 - 117.1 = I (1.97/0.225^2)
35.5 = I (38.9)
0.912 kg*m^2 = I


Thank you :)
 
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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