Equilibrium — Question 1

Section 12-1 · Physics pages 1772–1773

Question 1 Rest at one instant is not enough to establish equilibrium A 2 kg block lies at rest at t = 0 on a frictionless horizontal surface. A constant horizontal force 6 N begins acting at that instant. Gravity and the normal cancel. 2 kg 6 N Schematic; labels, not drawing lengths, specify the data. (a) Is the block in equilibrium at t = 0? Find its velocity after 0.20 s. (b) What additional observation, beyond an instantaneous photograph, would distinguish rest from sustained static equilibrium?
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Question 1 - Solution Rest at one instant is not enough to establish equilibrium 2 kg 6 N (a) Derivation and calculation After the force begins, the net horizontal force is nonzero: ∑ Fx = max = 6 N, ax = 6/2 = 3 m/s2 . With vx(0) = 0 and constant acceleration, vx(.20) = vx(0) + ax(.20) = .60 m/s . The block is instantaneously at rest but is not in equilibrium under this load. (b) Result and interpretation Sustained static equilibrium in the chosen inertial frame requires ∑⃗ F =⃗0, ∑⃗ τ =⃗0,⃗ vC(0) =⃗0,⃗ ω(0) =⃗0. A time sequence can test whether position and orientation remain unchanged under the same conditions. A photograph supplies only an instantaneous position, not a force or acceleration measurement. Physical check Here the displacement over .20 s is 1 2(3)(.20)2 = .060 m, demonstrating departure from rest.

Question 1

Rest at one instant is not enough to establish equilibrium

A 2kg2\,\mathrm{kg} block lies at rest at t=0t=0 on a frictionless horizontal surface. A constant horizontal force 6N6\,\mathrm N begins acting at that instant. Gravity and the normal cancel.

Diagram shown in the original SVG worksheet.

  1. Is the block in equilibrium at t=0t=0? Find its velocity after 0.20s0.20\,\mathrm s.

  2. What additional observation, beyond an instantaneous photograph, would distinguish rest from sustained static equilibrium?

Question 1 - Solution

Rest at one instant is not enough to establish equilibrium

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

After the force begins, the net horizontal force is nonzero: ∑Fx=max=6N,ax=6/2=3m/s2.\sum F_x=ma_x=6\ \mathrm N,\qquad \boxed{a_x=6/2=3\ \mathrm{m/s^2}}. With vx(0)=0v_x(0)=0 and constant acceleration, vx(.20)=vx(0)+ax(.20)=.60m/s.\boxed{v_x(.20)=v_x(0)+a_x(.20)=.60\ \mathrm{m/s}}. The block is instantaneously at rest but is not in equilibrium under this load.

(b) Result and interpretation

Sustained static equilibrium in the chosen inertial frame requires ∑F→=0→,∑τ→=0→,v→C(0)=0→,ω→(0)=0→.\sum\vec F=\vec0,\qquad\sum\vec\tau=\vec0,\qquad \vec v_C(0)=\vec0,\quad\vec\omega(0)=\vec0. A time sequence can test whether position and orientation remain unchanged under the same conditions. A photograph supplies only an instantaneous position, not a force or acceleration measurement.

Physical check

Here the displacement over .20.20 s is 12(3)(.20)2=.060\tfrac12(3)(.20)^2=.060 m, demonstrating departure from rest.