Pulley system on an inclined plane with friction.

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SUMMARY

The discussion centers on calculating the coefficient of kinetic friction (µk) for a pulley system involving two blocks on an inclined plane, with an acceleration of 4.1 m/s² and a velocity of 1.9 m/s. The user initially calculated µk to be approximately 1.41 but received feedback indicating errors in the equations used. The correct approach involves applying the equation ƩF = (m1 + m2)a, emphasizing that the system is accelerating rather than moving at a constant velocity. The user is encouraged to reassess their calculations based on this clarification.

PREREQUISITES
  • Understanding of Newton's Second Law (ƩF = MA)
  • Knowledge of forces acting on inclined planes
  • Familiarity with the concept of kinetic friction (Ff = µk * Fn)
  • Basic algebra for solving equations
NEXT STEPS
  • Review the derivation of forces on inclined planes with friction
  • Learn about the dynamics of pulley systems in physics
  • Study the relationship between acceleration, mass, and net force
  • Practice solving problems involving kinetic friction and inclined planes
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of friction and pulley systems in action.

cmcraes
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Homework Statement



Two blocks are connected by a thin inextensible string over a frictionless massless pulley
as shown on the picture below
The acceleration of gravity is 9.8 m/s2

Given that the two blocks move at 1.9 m/s
under an acceleration of 4.1 m/s2,
calculate the coefficient µk of kinetic friction between
the left block and the incline.

Homework Equations

*
ƩF=MA
Ffk*Fn
m2-T = m2*a

*(any and all derived equations as well)


The Attempt at a Solution


Tension minus parallel force of gravity minus force of friction equals zero because its traveling at a constant velocity.

(110kg*(9.8m/s2-4.9m/s2)) - (37kg*9.8m/s2*sin(33)) - (μ*9.8m/s2*37kg*cos(33))

after doing the arithmetic and algebra i got to:
429.5138859N = (μ*9.8m/s2*37kg*cos(33))

Dividing both sides by (9.8m/s2*37kg*cos(33))
i got μ=1.412400969

I got this incorrect twice before and know have noticed my previous errors and fixed them, but was wondering if anyone can verify with me that this is the right/ wrong answer? thanks!
 

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cmcraes said:

Homework Statement



Two blocks are connected by a thin inextensible string over a frictionless massless pulley
as shown on the picture below
The acceleration of gravity is 9.8 m/s2

Given that the two blocks move at 1.9 m/s
under an acceleration of 4.1 m/s2,
calculate the coefficient µk of kinetic friction between
the left block and the incline.

Homework Equations

*
ƩF=MA
Ffk*Fn
m2-T = m2*a

*(any and all derived equations as well)


The Attempt at a Solution


Tension minus parallel force of gravity minus force of friction equals zero because its traveling at a constant velocity.

The text of the problem is confusing, but the blocks do not move with constant velocity: There is acceleration of 4.1 m/s2. v=1.9 m/s at an instant.

cmcraes said:
(110kg*(9.8m/s2-4.9m/s2)) - (37kg*9.8m/s2*sin(33)) - (μ*9.8m/s2*37kg*cos(33))


The equation is not correct. It should be ƩF=(m1+m2)a, the sum of all external forces is equal to the acceleration of the whole system multiplied by the mass of the whole system. What is that 4.9 m/s2?

ehild
 

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