- #61
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That is the same as the general equation I obtained.Jpyhsics said:
Finding the acceleration of two masses on a pulley system
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SUMMARY
The discussion focuses on calculating the acceleration of two masses, m1 (7.40 kg) and m2 (3.70 kg), connected by a massless pulley system. The system is initially in equilibrium but begins to slide when the surface accelerates to the right at 5.43 m/s². Key insights include that the horizontal acceleration of m2 is zero, while the frictional force acting on m1 is determined by the coefficient of friction, which is equal for static and kinetic states (µs=µk). The relationship between the accelerations of m1 and m2 is established as a1 = (1/2)(a2), where a2 is the acceleration of m2.
PREREQUISITES- Understanding of Newton's second law (F=ma)
- Knowledge of pulley systems and tension forces
- Familiarity with friction coefficients (static and kinetic)
- Ability to analyze free body diagrams
- Study the effects of friction on pulley systems in detail
- Learn about free body diagram techniques for complex systems
- Explore the implications of constant string length in pulley dynamics
- Investigate the relationship between acceleration and tension in connected masses
Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to pulley systems and acceleration calculations.
- #62
- #63
FriedChicken885
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haruspex said:
Okay, thank you very very very much for all of your help!
- #64
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Should one not consider that the tension supporting the hanging mass is no longer parallel to the vertical side of the table but at an angle?
Also, I think that ##g=9.81 ~m/s^2## in the expression posted by OP in #62 should be replaced by an effective acceleration. This problem can be solved quite simply in the accelerated frame with additional fictitious horizontal forces ##-m_i\vec a_0## acting on masses ##m_i##. Here, ##\vec a_0## is the given horizontal acceleration of the table.
Also, I think that ##g=9.81 ~m/s^2## in the expression posted by OP in #62 should be replaced by an effective acceleration. This problem can be solved quite simply in the accelerated frame with additional fictitious horizontal forces ##-m_i\vec a_0## acting on masses ##m_i##. Here, ##\vec a_0## is the given horizontal acceleration of the table.
- #65
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we are asked for the accelerations immediately after the surface has started to accelerate. The string is still vertical.kuruman said:
I don't see why. Do you get a different answer? If so, please post your solution.kuruman said:
- #66
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I don't get a different answer, but I am bothered by the fact that one writes the same equation for the hanging mass whether the pulley supporting it accelerates or not. When the pulley accelerates, the hanging mass must have a horizontal acceleration at t = 0 because its horizontal position changes a moment later. Should this not affect the tension? I need to think about this some more.
On Edit: OK, I got it. At t = 0 the horizontal acceleration is instantaneously zero and then changes as it reaches a constant value in the steady state.
On Edit: OK, I got it. At t = 0 the horizontal acceleration is instantaneously zero and then changes as it reaches a constant value in the steady state.
Last edited:
- #67
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Right.kuruman said:
Or maybe oscillates.kuruman said:
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