Applying Newton's Laws — Question 1

Question 1

Two unequal hanging masses share one acceleration magnitude

Take $g=10$ m/s$^2$. Masses 3 kg and 5 kg hang from the ends of a light inextensible cord over an ideal fixed pulley. The cord stays taut; neglect pulley inertia and resistance. They are released from rest.

1. Find acceleration and tension.

2. Find the speed after the heavier mass descends 0.8 m.

3. Check the tension against the two weights and explain the result.

Question 1 - Solution

Two unequal hanging masses share one acceleration magnitude

(a) Write one equation for each mass.

$$50-T=5a,\quad T-30=3a\ \Rightarrow\ \boxed{a=\frac{20}{8}=2.5\,\mathrm{m/s^2},\quad T=37.5\,\mathrm N}$$ For the 5 kg side downward positive, $50-T=5a$. For the 3 kg side upward positive, $T-30=3a$. Adding gives $a=20/8=2.5$ m/s$^2$, then $T=37.5$ N.

(b) Use the constant acceleration.

$$v^2=v_0^2+2a\Delta y=2(2.5)(0.8)=4\ \Rightarrow\ v=2\,\mathrm{m/s}$$ $v^2=2a\Delta y=2(2.5)(0.8)=4$, so both speeds are 2 m/s while the cord remains taut.

(c) Test the signs physically.

$$T-30=7.5>0,\quad50-T=12.5>0\ \Rightarrow\ 30<T<50\,\mathrm N$$ $30<T<50$ N. Tension exceeds the lighter weight to accelerate it upward and is smaller than the heavier weight to let that mass accelerate downward. It is not equal to either weight during acceleration.