Block of mass Rotational Problem

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SUMMARY

The discussion centers on a physics problem involving two blocks connected by a massless string over a disk-shaped pulley. The blocks have masses of m1 = 2.00 kg and m2 = 6.00 kg, with a pulley mass of M = 10.0 kg and radius R = 0.250 m. The coefficient of kinetic friction is 0.360. The key equations for solving the problem include torque equations and the relationship between tensions T1 and T2. The confusion arises from the sign convention used for the tensions, where clockwise torque is considered negative.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and torque
  • Knowledge of free-body diagrams
  • Basic principles of friction in physics
NEXT STEPS
  • Study the concept of torque and its applications in rotational motion
  • Learn about free-body diagram construction for complex systems
  • Explore the relationship between tension and acceleration in pulley systems
  • Investigate the effects of friction on motion in inclined planes
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Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators looking for examples of complex pulley systems.

AriAstronomer
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Homework Statement


A block of mass m1 = 2.00 kg and a block of mass m 2 =
6.00 kg are connected by a massless string over a pulley
in the shape of a disk having radius R = 0.250 m and
mass M = 10.0 kg. These blocks are allowed to move on
a fixed block – wedge of angle 30.0°, as shown in
Figure P10.39. The coefficient of kinetic friction for
both blocks is 0.360. Draw free-body diagrams of both
blocks and of the pulley. Determine (a) the acceleration
of the two blocks and (b) the tensions in the string on
both sides of the pulley.

Diagram: http://www.webassign.net/pse/pse6_p10-37.gif

Homework Equations





The Attempt at a Solution


Now I know how to solve this problem, the only thing I'm stuck on, is the pulley. I know that you have to make a substitution for the acceleration using:
Torque = (T1 + T2)R = I(alpha)
T1 + T2 = I(a/R^2).
The only thing I don't understand is why T1 is negative and T2 is positive (T1 being the tension affecting m1, and T2 affecting T2). By R.H.R, torque1 = T1xR = out of page, which I just assumed should be positive, and thus T2 should be negative, but I only get the right answer if I have T2 - T1 = I(a/R^2), and not T1 - T2 = I(a/R^2). I don't think I'm missing a sign anywhere else, this is kind of a popular problem, and have compared my solutions to others.

Any help would be appreciated.
Ari
 
Physics news on Phys.org
nvm. Clockwise = negative
 

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