## Related questions with answers

A uniform rope of cross-sectional area 0.50 cm² breaks when the tensile stress in it reaches

$6.00 × 10^6 N/m^2.$

(a) What is the maximum load that can be lifted slowly at a constant speed by the rope? (b) What is the maximum load that can be lifted by the rope with an acceleration of 4.00 m/s²?

Solution

Verified$\begin{align*} \sigma_\text{Ta}&=\dfrac{F_{\bot\text{a}}}{A} \\ \implies F_{\bot \text{a}} &= \sigma_\text{Ta} A \\ F_{\bot \text{a}} &= m_\text{a}g \\ \implies m_\text{a}&= \dfrac{\sigma_\text{Ta} A}{g} \\ &= \dfrac{\left(6.00 \cdot 10^6\right) \left(0.50 \cdot 10^{-4}\right)}{9.8} \\ &=30.6 \text{ kg} \end{align*}$

First, we write the equation for the definition of tensile strength ($\sigma_\text{T}$) and we solve for the tensile force $F_{\bot\text{a}}$. In this occasion, since the load is being moved at a constant speed, the force $F_{\bot\text{a}}$ is equal to the weight of the load. Therefore, we can substitute $F_{\bot\text{a}}$ for $m_\text{a}g$ in the first equation. Finally, we solve for the mass $m_\text{a}$.

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