Dynamics: Minimum Force Problem

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SUMMARY

The discussion centers on solving the Minimum Force Problem in physics, specifically determining the smallest mass (M) required for a system to remain at rest on an incline with a static friction coefficient of 0.100. The correct answer is established as 4.98 kg. Key equations utilized include Fnet = ma, the sum of forces in both horizontal and vertical directions, and the frictional force equation Ff = μ・FN. The user JM92 clarified the placement of forces in their diagram, leading to the correct solution.

PREREQUISITES
  • Understanding of static friction and its coefficient
  • Familiarity with Newton's laws of motion
  • Knowledge of vector components in physics
  • Ability to analyze free-body diagrams
NEXT STEPS
  • Study the principles of static and kinetic friction in detail
  • Learn how to construct and analyze free-body diagrams
  • Explore applications of Newton's laws in various physics problems
  • Investigate the effects of incline angles on force calculations
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This discussion is beneficial for physics students, educators, and anyone interested in mastering concepts related to forces, friction, and equilibrium in mechanics.

JM92
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Hi, nice to meet everybody here at physicsforums.com! I'm JM92 and I have a physics problem I need help with.

Homework Statement



Find the smallest value of the mass M for which the system will remain at rest given that the coefficient of static friction is 0.100
[PLAIN]http://img695.imageshack.us/img695/6328/physics2.jpg

Homework Equations



Fnet = ma
∑F(Horizontal) = 0
∑F(Vertical) = 0
Ff = μ・FN
Fgx = Fg・sinθ
Fgy = Fg・cosθ

The Attempt at a Solution


[PLAIN]http://img3.imageshack.us/img3/4139/physics11.jpg
I retyped and redrew everything with MS paint.. Hopefully that made it easier to read and not harder..:smile:

PS. The correct answer is 4.98kg.

Thank you for your help!

EDIT: Sorry, the thread title should be called "Minimum Mass Problem"
 
Last edited by a moderator:
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You've drawn Ff in the wrong place. M is the mass of the second block; if it is any smaller, the first block will slide down the incline.

How small can M be?
 
vertigo said:
You've drawn Ff in the wrong place. M is the mass of the second block; if it is any smaller, the first block will slide down the incline.

How small can M be?

Ohh! I was looking at it backwards, then. With Ff in the opposite direction, I got the correct answer. Thank you very much!
 

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