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A 76.0-kg person is being pulled away from a burning
Chapter 4, Problem 42(choose chapter or problem)
A 76.0-kg person is being pulled away from a burning building as shown in Figure 4.41. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.
Figure 4.41 The force \(\mathrm{T}_{2}\) needed to hold steady the person being rescued from the fire is less than her weight and less than the force \(\mathrm{T}_{1}\) in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force).
Questions & Answers
QUESTION:
A 76.0-kg person is being pulled away from a burning building as shown in Figure 4.41. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.
Figure 4.41 The force \(\mathrm{T}_{2}\) needed to hold steady the person being rescued from the fire is less than her weight and less than the force \(\mathrm{T}_{1}\) in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force).
ANSWER:
Step 1 of 2
The tension in the two ropes can be calculated from the free body diagram as shown in the figure below.
Since the person is momentarily motionless ,tension in the string is in static equilibrium. So,
the sum of all forces acting in x and y direction are zero.
\(\Sigma x=0\) and \(\Sigma y=0\)
Therefore, \(\Sigma x=0=-T_{1} \cos 75^{\circ}+T_{2} \cos 10^{\circ}\)
\(\Sigma y=0=T_{1} \sin 75^{\circ}+T_{2} \sin 10^{\circ}-W\)
where, \(\mathrm{T}_{1}\) and \(\mathrm{T}_{1}\) are tension in the two ropes.
W = Weight of the person = m g
\(\mathrm{m}=75 \mathrm{~kg}\)
\(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\)