Applying Newton's Laws — Question 9

Question 9

A movable pulley changes the acceleration constraint

Take $g=10$ m/s$^2$. One end of a light cord is fixed to a ceiling. It passes under an ideal massless movable pulley carrying a 6 kg load, then over an ideal fixed pulley to a free hanging 2 kg mass. All relevant cord segments are vertical and taut. Ignore all pulley inertia and resistance.

1. With downward coordinates y for the 6 kg load and z for the 2 kg mass, derive the acceleration constraint.

2. Find both accelerations and the tension.

3. If the load descends 0.10 m from rest, find how far and how fast the 2 kg mass moves.

Question 9 - Solution

A movable pulley changes the acceleration constraint

(a) Count the changing cord segments.

$$L=2y+z+L_0\ \Rightarrow\ 2v_y+v_z=0\ \Rightarrow\ a_z=-2a_y$$ Two segments lengthen when the movable pulley descends, so $2y+z=\text{constant}$. Thus $a_z=-2a_y$ and similarly $v_z=-2v_y$.

(b) Use two tensions on the movable load.

$$60-2T=6a_y,\quad20-T=-4a_y\ \Rightarrow\ \boxed{a_y=10/7,\ a_z=-20/7,\ T=180/7}\ \text{(SI)}$$ $60-2T=6a_y$ and $20-T=2a_z=-4a_y$. Hence $T=20+4a_y$, so $20=14a_y$. The load accelerates downward at $10/7$ m/s$^2$, the small mass upward at $20/7$ m/s$^2$, and $T=180/7$ N.

(c) Apply the same rope ratio to displacement and speed.

$$\Delta z=-2(0.10)=-0.20\,\mathrm m,\quad |v_z|=2\sqrt{2a_y\Delta y}=2\sqrt{2/7}=1.069\,\mathrm{m/s}$$ The small mass rises 0.20 m. The load speed is $\sqrt{2(10/7)(0.10)}=\sqrt{2/7}$ m/s, so the small mass speed is $2\sqrt{2/7}\approx1.069$ m/s. Equal tension does not imply equal acceleration magnitudes in this geometry.