Two blocks connected by string passing over a pulley on an incline (1 Viewer)

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1. The problem statement, all variables and given/known data

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.

2. Relevant equations



3. 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!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 

rl.bhat

Homework Helper
4,433
5
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:
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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 :)
 
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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?
 

rl.bhat

Homework Helper
4,433
5
Sorry.

T2 - T1 = I*α/r. = I*a/r^2
 
38
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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 :)
 

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