Some Particular Forces — Question 7

Question 7

Equal tensions in an ideal rope rely on the connector model

A straight vertical rope segment is pulled upward at its top with force $T_t$ and downward at its bottom with force $T_b$. It accelerates upward at 2 m/s$^2$. Take $g=10$ m/s$^2$.

1. For a massless ideal segment, relate the two tensions.

2. For a 0.5 kg segment, find $T_t-T_b$.

3. If $T_b=20$ N in the massive case, find $T_t$ and explain why equality fails.

Question 7 - Solution

Equal tensions in an ideal rope rely on the connector model

(a) Apply the zero-mass idealization.

$$T_t-T_b-m_rg=m_ra,\quad m_r=0\ \Rightarrow\ T_t=T_b$$ $T_t-T_b=0$ for the massless segment, hence equal tensions. Its modeled weight is also zero.

(b) Include inertia and weight.

$$T_t-T_b=m_r(g+a)=0.5(10+2)=6\,\mathrm N$$ $T_t-T_b-m_rg=m_ra$ gives $T_t-T_b=m_r(g+a)=0.5(12)=6$ N.

(c) Supply the rope’s own demand.

$$\boxed{T_t=20+6=26\,\mathrm N},\quad6=5+1\quad\mathrm N$$ $T_t=26$ N. The extra 6 N both supports the rope’s 5 N weight and provides its 1 N upward resultant. Equal tension is an ideal-connector result, not an unconditional rule for every rope.