Physics equations/Static forces: Difference between revisions

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imported>Guy vandegrift
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Latest revision as of 21:53, 20 June 2023

Most applications of Newton's laws require only a few equations

  • ma=FjFij=Fji

   Fx=FcosθFy=Fsinθ       Fx2+Fy2=F2tanθ=sinθcosθ <section end=NewtonsThreeLaws/>

Friction and the normal force

  • fk=μkN is the force friction when an object is sliding on a surface, where μk ("mew-sub-k") is the kinetic coefficient of friction, and N is the normal force.
  • fsμsN establishes the maximum possible friction (called static friction) that can occur before the object begins to slide. Usually μs>μk.

Also, air drag often depends on speed, an effect this model fails to capture. <section end=FrictionKineticStatic/> These equations for static and kinetic friction almost always are valid only as approximations.

*** Problem: Generating equations from adding forces
Three forces of tension are acting on the small grey circle at the center.
Find the x and y components of the three forces on the small grey circle at the center

Template:Hidden begin Solution:

T1x=T1cosθ1 ,        T1y=T1sinθ1
T2x=0 ,                             T2y=mg
T3x=T3cosθ3 ,          T3y=T3sinθ3

Template:Hidden end

*** Problem: Almost proving Netwon's third law (one dimensional)
Two carts are connected by a string; a force is applied to one. By Newton's third law, the force on 1 by 2 equals the force on 2 by t. But we do not need Newton's third law to solve this problem.
Consider two carts of mass known masses (m1 and m2) connected by a taught string, with one mass (m2) experiencing and external force of known magnitude. Assume that the collection of objects held together by a taught string obeys F=(m1+m2)a where F is the sum of all external forces. But we do not assume that string (in the middle) exerts equal and opposite forces on the two objects. (In this way we prove Newton's third law.)

Template:Hidden begin Solution: We take the acceleration, as well as the two internal forces as unknowns (we have three unknowns because we make no assumption about the equality of the two internal forces transmitted by the string, which we assume to have negligible mass. With three unknowns, we seek three equations. Beginning with Newton's second law (ΣF=ma) applied to and M=m1+m2, we have:

  1. Fext =(m1+m2)a
  2. F21 =m1a
  3. ΣF = Fext + F12 = m2a

(They all have the same acceleration because the string is assumed taught.)

  • The student should be able to show that a=Fext/(m1+m2) and that F21=F12=m1Fextm1+m2. And, the student should be able to show this without ever making the assumption that F21=F12 .

Template:Hidden end

*** Problem: A free-body diagram on a system with many parts
Power Pulley
Make free body diagrams for each component of this system. Include attachments to the ceiling.

Template:Hidden begin Solution:

Power Pulley

Template:Hidden end

*** Problem: Motion under the influence of kinetic friction
Let this be an object in motion, acting under friction, the normal force, and gravity.
If this object is in motion and the coefficient of kinetic friction is μ, find the acceleration.

Template:Hidden begin Solution: This problem is unsolved. Template:Hidden end

[More problems]