Solution for Question 47
Analysis of Sliding Condition
When the block begins to slide, the horizontal tension $T_0$ exerted by the rope must equal the maximum limiting friction force of the block.
$$ T_{0,new} = f_{max} = \mu M g = 37.5 \text{ N} $$Recalculating Rope Mechanics
We need to find the new height $h$ of the upper end. The properties of the rope ($\lambda g = 5$ N/m, $W = 20$ N) remain unchanged.
Using the vector relationship for the rope tensions again:
$$ T_{top}^2 = T_{0,new}^2 + W^2 $$ $$ T_{top}^2 = (37.5)^2 + (20)^2 $$Let’s simplify the math. Note that $37.5 = \frac{75}{2}$ and $20 = \frac{40}{2}$.
$$ T_{top} = \frac{1}{2} \sqrt{75^2 + 40^2} $$Divide by common factor 5 inside the square root ($15^2 + 8^2$):
$$ T_{top} = \frac{5}{2} \sqrt{15^2 + 8^2} = \frac{5}{2} \sqrt{225 + 64} = \frac{5}{2} \sqrt{289} $$ $$ T_{top} = \frac{5}{2} \times 17 = \frac{85}{2} = 42.5 \text{ N} $$Calculate Height
Using the catenary relation derived in Question 45:
$$ h = \frac{1}{\lambda g}(T_{top} – T_{0,new}) $$ $$ h = \frac{1}{5}(42.5 – 37.5) $$ $$ h = \frac{1}{5}(5.0) = 1.0 \text{ m} $$Answer: (c) 1.0 m