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Force P to accelerate 600 N block

Short answer

Vertical: N = 600 – 0.342P. Horizontal: 0.9397P – 0.25(600 – 0.342P) = (600/9.81)×2, giving P = 265.6 N. Dynamic equilibrium: include an inertia force ma opposite acceleration, then ΣF = 0.

A 600 N block rests on a horizontal plane. Find the magnitude of P required to produce the block an acceleration of 2 m/s² to the right. The coefficient of friction is 0.25. What do you mean by dynamic equilibrium? [Figure: block of 600 N pulled by force P acting at 20° above the horizontal.]

Reading of the figure

The 600 N block is on a horizontal plane. P pulls the block at 20° above the horizontal, so its vertical component P sin 20° lifts the block slightly and the normal reaction is N = W − P sin 20°. μ = 0.25 and a = 2 m/s² to the right.

          P
         /
        / 20°
   ____/
  |600N|--> a = 2 m/s²
 ------------------

Equations of motion

  • Vertical: N = 600 − P sin 20° = 600 − 0.342P
  • Horizontal: P cos 20° − μN = (W/g) a P(0.9397) − 0.25(600 − 0.342P) = (600/9.81)(2) 0.9397P − 150 + 0.0855P = 122.32 1.0252P = 272.32

P = 265.6 N

(Check: N = 600 − 0.342(265.6) = 509.2 N, friction = 0.25 × 509.2 = 127.3 N, and 265.6 cos 20° − 127.3 = 122.3 N = (600/9.81)(2) ✓.)

Dynamic equilibrium

A body accelerating under a resultant force can be treated as if it were in equilibrium (D'Alembert's principle) by adding an imaginary inertia force equal to m a, directed opposite to the acceleration: ΣF − m a = 0 Here the inertia force = (600/9.81)(2) = 122.3 N acts to the left on the block, and the usual equations of statics (ΣFx = 0, ΣFy = 0, ΣM = 0) can then be applied. The block is then said to be in dynamic equilibrium.

✓ Answer checked by Claude Opus 5.5 (high effort)

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