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S2.8 Frictional force

[A] About the force

Commentary

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Commentary

  • at low enough FA, have zero a
  • there is static equilibrium
  • hence FF=FA
Graphic - No title - frictionone
  • at big enough FA, have non-zero a
  • hence FF<FA
Graphic - No title - frictiontwo
 

[B] Example problem

A block of mass m=2 kg rests on a plane inclined at an angle θ=45 to the horizontal. A force $\vec{F}_A$ of magnitude 30 N is applied to it, horizontally. The block moves up the plane with constant velocity.
Graphic - No title - blockgoingup

Write down equations for (a) the normal contact force between box and plane and (b) the net force acting up the plane. Deduce the value of the coefficient of kinetic friction.

Solution

Solution

  • First things

    • choose system: the mass
    • identify forces: FG, FN and FF
    • draw free body diagram
    • choose coordinates
    Graphic - No title - blockgoinguptwo
  • Apply 2nd law

    Block moves at constant velocity; so net force on it must be zero.
    • perpendicular to plane (y-direction):

      0=may=FAsinθ+mgcosθ-FN

    • parallel to plane (x-direction):

      0=max=FAcosθ-mgsinθ-FF

  • Invoke knowledge of forces

    • During slipping: FF=μkFN
    • Hence
      \[ \mu_k = \frac{F_F}{F_N} = \frac{F_A\cos{\theta} -mg\sin \theta}{ F_A\sin{\theta} +mg\cos \theta} \]
 

Results

\[ \mu_k = \frac{F_F}{F_N} = \frac{F_A\cos{\theta} -mg\sin \theta}{ F_A\sin{\theta} +mg\cos \theta} = \frac{30-20}{30+20} =0.2 \]

Check It!
TFriction Between Box and Rug
MBlock on a Frictionless Plane

Learning Resources

Textbook: HRW Chapter 6.2-6.3
Course Questions:
Self-Test Questions: