Blocks on Inclined Ramp Connected by Pulley

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SUMMARY

The discussion focuses on a physics problem involving two blocks of equal mass connected by a rope over a frictionless pulley on a 30-degree inclined ramp. To keep the upper block stationary, the required coefficient of static friction was calculated, resulting in an incorrect negative value for static friction. For the second part, the velocity of the blocks when the lower mass descends by 0.5 meters was determined using conservation of energy, yielding a speed of 3.13 m/s. The analysis emphasizes the importance of correctly applying force equations and energy conservation principles in solving such problems.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static friction and its coefficient
  • Familiarity with conservation of energy principles
  • Ability to analyze forces on inclined planes
NEXT STEPS
  • Study the derivation of the coefficient of static friction in inclined plane problems
  • Learn about energy conservation in mechanical systems
  • Explore the dynamics of systems involving pulleys and multiple masses
  • Review the application of free-body diagrams in physics problems
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to forces, friction, and energy conservation in dynamic systems.

chickyroger
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Homework Statement


Two blocks of equal mass are connected by a rope of negligible mass that is passed over a frictionless pully. The angle made by the horizontal plane and the ramp is 30 degrees.

852c3d73-7b60-4e87-b7c5-d6cc1282dc10.png


a) If neither block moves, what is the required coefficient of static friction to keep the upper block from moving?
b) If the ramp is replaced by a frictionless ramp, how fast will the blocks be traveling when the lower mass descends by 0.5 meters?

Homework Equations


F = ma
KEinitial + PEinitial = KEfinal + PEfinal

The Attempt at a Solution


a) For the block hanging on the pulley:
Fnet = Mg - FT
ma = Mg - FT

For the block on the ramp:
ma = FT - Mgsin30 - [itex]\mu_{k}[/itex]gcos30

Then I set the two equations equal to each other:
Mg - FT = FT - Mgsin30 - [itex]\mu_{k}[/itex]gcos30
Mg = -Mg(sin30 + [itex]\mu_{k}[/itex]cos30)
-1 = sin30 + [itex]\mu_{k}[/itex]cos30
[-1-sin30]/cos30 = [itex]\mu_{k}[/itex] = -[itex]\sqrt{3}[/itex]

I got a negative number for static fricton?

b) I have no idea how to do this problem ): Do I use the conservation of energy formula to find velocity? so it'd be something like:
0 + Mg(0.50) = 1/2Mv2 + 0
v = 3.13 m/s

Someone please check if I am right!
 
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I think its better to work without the energy way, let's use only the forces.

If we make the x' axe as the direction of the ramp and the y' axe perpendicular to it, wee see that the forces in the x' axes are:
[itex]P[/itex]
[itex]Psen30º[/itex]
and Friction force [itex]Fa = N.u = P.cos30º.u[/itex]


Equaling:



[itex]1,5P = P.cos30º.u -> u = \sqrt{3}[/itex]

Without the friction force the forces in the x' axe are [itex]1,5P = 2ma -> a=\sqrt{30}/2[/itex]

[itex]V² = Vo² + 2a\Delta S -> V = \sqrt{2a\Delta S} = \sqrt[4]{7,5}[/itex]
 
For part a

1. When the system is in equilibrium, ma = ?
2. In what direction does FT act on the mass on the incline? In what direction does mgsin30 act on that mass?

For part b

Yes you can use conservation of energy, but you need to determine the initial and final PE and KE of both blocks together, Each has initial speed 0 and final speed V. For PE of each, initial and final, you need to define a reference plane. Assume, at the start, the inclined block is 0.5 m up the slope from the bottom of the incline, and the hanging block is at the bottom of the incline. For the final position, the inclined block will be at the bottom of the incline, and the hanging block will be 0.5 m down from the bottom of the incline.
 

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