Some Particular Forces — Question 6

Question 6

An ideal taut cord pulls along its own line

A 3 kg mass hangs from a light taut cord. The mass accelerates upward at 2 m/s$^2$. Take $g=10$ m/s$^2$.

1. Find the tension.

2. Find tension for downward acceleration of 2 m/s$^2$.

3. Can the same cord provide a downward force if the mass requires one?

Question 6 - Solution

An ideal taut cord pulls along its own line

(a) Balance force difference with inertia.

$$T-mg=ma_y\ \Rightarrow\ \boxed{T=3(10+2)=36\,\mathrm N}$$ $T-mg=ma$ gives $T=3(10+2)=36$ N.

(b) Reverse acceleration, not the tension direction.

$$T=3(10-2)=24\,\mathrm N,\quad T-mg=-6\,\mathrm N$$ $T=3(10-2)=24$ N. The tension still pulls upward even though the net force is downward.

(c) Use the unilateral nature of a cord.

$$T\ge0\ \Rightarrow\ a_y\ge-g\quad\text{for this taut-cord inventory}$$ An ordinary flexible cord under tension pulls; it does not push the hanging mass downward. If the equations require negative tension, the assumed taut configuration fails and a slack-cord or additional-interaction model is needed.