Two blocks and a pulley, friction

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SUMMARY

The discussion centers on a physics problem involving two masses connected by a string over a frictionless, massless pulley. The hanging mass is 4.0 kg, while the mass on the horizontal surface is 6.0 kg, with a coefficient of kinetic friction of 0.2. The key equation used is T = mBg - mBa, but the user struggles to incorporate friction into their calculations for acceleration. The correct approach requires considering the effects of friction on the tension and acceleration of both masses.

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  • Understanding of Newton's laws of motion
  • Familiarity with tension in strings and pulleys
  • Knowledge of kinetic friction and its effects on motion
  • Ability to solve equations involving multiple masses
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  • Study the equations of motion for connected objects
  • Explore the concept of net force and its application in friction scenarios
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of pulley systems and friction in action.

nbaumann
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Homework Statement



Two masses are connected by a string which passes over a frictionless, massless pulley. One mass hangs vertically and one slides on a horizontal surface. The horizontal surface has a coefficient of kinetic friction 0.2. The vertically hanging mass is 4.0 kg and the mass on the horizontal surface is 6.0 kg. Find the acceleration of the mass.

Homework Equations



T=mBg-mBa

The Attempt at a Solution



I've basically done this problem without factoring friction, and this my problem, I don't know how to factor friction.

a = (mBg)/(mA + mB) is what I used.


Thanks!
 
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nbaumann said:

Homework Statement



Two masses are connected by a string which passes over a frictionless, massless pulley. One mass hangs vertically and one slides on a horizontal surface. The horizontal surface has a coefficient of kinetic friction 0.2. The vertically hanging mass is 4.0 kg and the mass on the horizontal surface is 6.0 kg. Find the acceleration of the mass.

Homework Equations



T=mBg-mBa

The Attempt at a Solution



I've basically done this problem without factoring friction, and this my problem, I don't know how to factor friction.

a = (mBg)/(mA + mB) is what I used.


Thanks!
Welcome to Physics Forums,

You should look again at the tension in the string, is that the only condition that the tension needs to satisfy? What about the other block?
 
Thanks for your reply.

This was really as far as I got. Wouldn't the tension just apply once? As there is only one string?
 

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