Question

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

6.00×106N/m2.6.00 × 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

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σTa=FaA    Fa=σTaAFa=mag    ma=σTaAg=(6.00106)(0.50104)9.8=30.6 kg\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 (σT\sigma_\text{T}) and we solve for the tensile force FaF_{\bot\text{a}}. In this occasion, since the load is being moved at a constant speed, the force FaF_{\bot\text{a}} is equal to the weight of the load. Therefore, we can substitute FaF_{\bot\text{a}} for magm_\text{a}g in the first equation. Finally, we solve for the mass mam_\text{a}.

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